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131 Solved Past Papers
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Electrostatics Past Papers
Solved past paper MCQs for Electrostatics from official UHS, NUMS, SZABMU, DUHS, and KMU examinations. Includes verified distractor autopsies and step-by-step cognitive explanations.
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Official printable A4 booklet containing all 131 verified past paper MCQs with unbroken options, 5-column rapid answer key, and Propolis cognitive error autopsies.
Q.178 The electric flash attachment to a camera contains a capacitor for storing the energy used to produce the flash. In one such unit, the potential difference between the plates of 20 F capacitor is 5 V. Calculate the energy that is used to produce the flash?
A
250 J
B
310 J
C
500 J
D
650 J
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
The total electrical potential energy stored in a charged capacitor can be found using its static capacitance and the applied voltage.
2. Formula:
$$ U = \frac{1}{2} C V^2 $$
3. Solution:
Given capacitance: \( C = 20 \text{ F} \).
Given voltage: \( V = 5 \text{ V} \).
Square the voltage: \( 5^2 = 25 \).
Substitute values: \( U = \frac{1}{2} (20) (25) \).
An electron-volt (eV) describes the work done (or kinetic energy gained) when a single electron is accelerated through a potential difference of exactly one volt.
2. Formula:
$$ W = q \Delta V \implies 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} $$
3. Solution:
Because it equates directly to a tiny fraction of a Joule, it is fundamentally a unit of Energy, heavily used in atomic and particle physics.
4. Why other options are incorrect:
Charge is measured in Coulombs. Current is measured in Amperes. Electric potential is measured in Volts. The 'volt' in the name often tricks students into guessing Option C.
Q.180 Which one of the following is the unit of electric field intensity?
A
Newton per Ampere
B
Newton per volt
C
Volt per Coulomb
D
Volt per meter
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
Electric field intensity can be measured either by the mechanical force it exerts on a charge or by the spatial gradient of its electric potential.
2. Formula:
$$ E = \frac{F}{q} = \frac{\Delta V}{\Delta r} $$
3. Solution:
Using the potential gradient definition (\( \Delta V / \Delta r \)), the unit naturally becomes Volts (for potential) divided by meters (for distance).
4. Why other options are incorrect:
Newton per Coulomb (N/C) is correct, but Newton per Volt or Ampere are dimensionally nonsensical for an electric field. Volt per Coulomb represents Energy per Charge squared.
Relative permittivity (often called the dielectric constant, \( \epsilon_r \)) is defined as the ratio of a material's absolute permittivity to the absolute permittivity of a perfect vacuum.
Because you are mathematically dividing a value by another value with the exact same SI units (\( \text{F/m} / \text{F/m} \)), the units perfectly cancel out.
This renders relative permittivity a pure, dimensionless scalar ratio.
4. Why other options are incorrect:
Option A is the correct unit for the absolute permittivity of free space (\( \epsilon_0 \)), not the relative ratio. Options B and C are incorrect mathematical variations.
Q.182 The electrostatic force between two point-charges is independent of one of the following quantities?
A
Distance between charges
B
Magnitude of charges
C
Medium between charges
D
Temperature of charges
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
Coulomb's Law provides an exact accounting of the variables required to calculate static electrical force.
2. Formula:
$$ F = \frac{1}{4 \pi \epsilon} \frac{q_1 q_2}{r^2} $$
3. Solution:
The equation explicitly requires: Charge magnitudes (\( q_1, q_2 \)), spatial distance (\( r \)), and the medium's permittivity (\( \epsilon \)).
Temperature does not appear in the foundational laws of macro-electrostatics and has no bearing on the isolated electrostatic force magnitude between two points.
4. Why other options are incorrect:
Distance, magnitude, and the intervening medium are the literal three pillars that define Coulomb's Law. Changing any of them will drastically alter the force.
The other options suffer from decimal placement errors (orders of magnitude). The value \( 6.25 \times 10^{18} \) is a universally memorized constant bridging microscopic particles to macroscopic current.
Q.185 What is the increase in force between two charges if the separation between them is decreased by 50 percent?
A
Becomes four times
B
Doubles
C
Increases by half
D
Triples
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
Electrostatic force scales according to the inverse square law of the distance separating the two charges.
2. Formula:
$$ F \propto \frac{1}{r^2} $$
3. Solution:
Decreasing the distance by 50 percent means the new distance is exactly half the original: \( r' = 0.5r = r/2 \).
Substitute this new distance into the denominator: \( F' \propto \frac{1}{(r/2)^2} \).
Square the term: \( F' \propto \frac{1}{r^2/4} \).
Invert the fraction: \( F' = 4 \left( \frac{1}{r^2} \right) \).
The final force is exactly 4 times the original force.
4. Why other options are incorrect:
Option B (doubling) occurs if you forget to square the distance change. Options C and D are not mathematically possible via the inverse square law for a 50% change.
Q.186 According to Coulomb's law, what happens to the electrostatic force between the 2-point charges if the distance between them is doubled?
A
The force becomes one-fourth
B
The force becomes half
C
The force doubles
D
The force remains the same
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
Moving charges further apart weakens their mutual electrostatic interaction following the inverse square law.
2. Formula:
$$ F \propto \frac{1}{r^2} $$
3. Solution:
Let the new distance be \( r' = 2r \).
Plug it into the proportion: \( F' \propto \frac{1}{(2r)^2} \).
Square the coefficient: \( F' \propto \frac{1}{4r^2} \).
The result is \( \frac{1}{4} \) of the original mathematical expression, proving the force drops to one-fourth.
4. Why other options are incorrect:
Option B assumes a linear \( 1/r \) drop-off. Option C falsely assumes force increases with distance. Option D assumes force is completely independent of distance.
Q.187 What does one Coulomb represent in terms of charges?
A
The amount of charge transported by a current of 1 Ampere in 1 second
B
The charge of 1 proton
C
The charge of 1 electron
D
The charge required to create a force of 1 Newton between two charges separated by 1 meter
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
The SI unit of charge (Coulomb) is actually a derived unit, officially defined through the macroscopic properties of flowing electric current (Amperes) and time.
2. Formula:
$$ Q = I \times t $$
3. Solution:
By standard SI definition, 1 Ampere represents 1 Coulomb of charge flowing past a specific point per second.
Therefore, multiplying 1 Ampere by 1 second inherently defines exactly 1 Coulomb.
4. Why other options are incorrect:
Options B and C refer to the elementary charge (\( 1.6 \times 10^{-19} \text{ C} \)). Option D describes an entirely different mathematical constraint (where \( F = k(1)^2/1^2 = 9 \times 10^9 \text{ N} \), not 1 N).
Q.188 Two points charges, \( +5\mu\text{C} \) and \( -5\mu\text{C} \) are placed at points A and B, respectively, which are separated by a distance 2d. What is the electric potential at the midpoint M of the line joining A and B?
A
\( 2kQ/d \)
B
\( kQ/d \)
C
\( -kQ/d \)
D
Zero
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
This setup describes a classic electric dipole. Electric potential is a scalar quantity, meaning the potentials from multiple charges simply add algebraically (keeping their positive or negative signs).
2. Formula:
$$ V_{net} = V_+ + V_- $$
3. Solution:
The total distance is \( 2d \), meaning the midpoint M is exactly distance \( d \) away from both charges.
Potential from positive charge: \( V_+ = \frac{k(+5\mu\text{C})}{d} \).
Potential from negative charge: \( V_- = \frac{k(-5\mu\text{C})}{d} \).
Options A, B, and C would apply if the question asked for Electric Field intensity (a vector, where the two fields would point in the exact same direction and constructively add together). Potential, being scalar, strictly cancels.
Q.189 In the case of a parallel plate capacitor, when the distance between the two plates is reduced to half and the area of the plate doubled, the capacitance
A
Increases four times
B
Increases six times
C
Is doubled
D
Remains the same
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
The capacitance of a parallel plate setup relies on the ratio of its surface area to its gap distance. Manipulating both variables simultaneously multiplies their respective effects.
2. Formula:
$$ C = \frac{\epsilon_0 A}{d} $$
3. Solution:
Let initial capacitance be \( C = \frac{\epsilon_0 A}{d} \).
The area is doubled: \( A' = 2A \).
The distance is halved: \( d' = d / 2 \).
Substitute into formula: \( C' = \frac{\epsilon_0 (2A)}{d/2} \).
The denominator's fraction flips up: \( C' = 2 \times 2 \left( \frac{\epsilon_0 A}{d} \right) = 4C \).
4. Why other options are incorrect:
Option D occurs if Area is halved and Distance is halved (canceling out). Option C occurs if only Area is doubled, or only Distance is halved. Combining them yields a 4x multiplier.
Q.190 If the dielectric material between the plates of the capacitor is removed, what happens to the electric field between the plates?
A
The electric field becomes zero
B
The electric field decreases
C
The electric field increases
D
The electric field remains the same
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
Assuming the charged capacitor is isolated (disconnected from a battery), removing the dielectric removes the internal opposing polarization field. Thus, the total net field rebounds to its original vacuum strength.
2. Formula:
$$ E_{vac} = E_{med} \times \epsilon_r $$
3. Solution:
While the dielectric was inserted, the electric field was suppressed (\( E_{med} = E_{vac} / \epsilon_r \)).
Removing the dielectric means \( \epsilon_r \) effectively returns to 1.
Because the trapped charge \( Q \) cannot escape, the lack of polarization mathematically forces the net electric field \( E \) to instantly increase back to its undamped vacuum state.
4. Why other options are incorrect:
Option B is exactly what happens when you insert the dielectric. Option D assumes the dielectric had no physical effect at all.
Q.191 Capacitance of a capacitor increases with decrease in:
A
Dielectric constant
B
Plate area
C
Permittivity
D
Plate separation
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
Capacitance relies on the geometric inverse proportionality between storage capability and the distance isolating the oppositely charged plates.
2. Formula:
$$ C = \frac{\epsilon A}{d} $$
3. Solution:
Because separation distance (\( d \)) resides strictly in the denominator, decreasing its value mathematically forces the total value of \( C \) to grow larger.
Physically, pushing the plates closer increases the mutual attraction between opposite charges, allowing the plates to hold more total charge at the same voltage.
4. Why other options are incorrect:
Decreasing Dielectric constant, Permittivity, or Plate Area (all numerator variables) would strictly cause the Capacitance to actively decrease.
An electron-volt is a localized unit representing the kinetic energy an electron gains when accelerated through a one-volt field.
2. Formula:
$$ W = q \Delta V $$
3. Solution:
Charge \( \times \) Voltage = Joules.
Since an electron-volt directly maps to \( 1.6 \times 10^{-19} \text{ Joules} \), it is inherently a measurement of Energy.
4. Why other options are incorrect:
Charge is Coulombs. Current is Amperes. Power is Watts (Energy over time). The presence of the word "Volt" often misleads students into associating it with potential rather than energy.
Q.193 Which one of the following is NOT a feature of electric forces?
A
They act on charges
B
They act on masses
C
They can be attractive
D
They can be repulsive
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
Electrostatic forces are a fundamental interaction strictly governed by the property of electric charge, entirely independent of the object's mass.
2. Formula:
$$ F_E = k \frac{q_1 q_2}{r^2} $$
3. Solution:
The formula for electric force dictates that objects must possess a non-zero charge (\( q \)) to interact electrostatically.
Gravitational forces act on masses. Electric forces inherently do not care about mass; a massive uncharged neutron feels exactly zero electric force.
4. Why other options are incorrect:
Options A, C, and D accurately describe the literal defining traits of electric forces: they require charges, and unlike gravity, they possess dual polarity (both attraction and repulsion).
Q.195 Electric intensity between two oppositely charge plates in the middle region is
A
Non-uniform
B
Uniform
C
Cannot be predicted
D
Variable
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
Deep inside the central gap of a parallel plate capacitor (ignoring the bulging 'fringing' effects at the far edges), the electric field lines run perfectly parallel and equally spaced.
2. Formula:
$$ E = \frac{V}{d} = \text{Constant} $$
3. Solution:
Because the field lines do not compress or expand in the central region, the electric intensity remains perfectly uniform at every geometric point between the plates.
4. Why other options are incorrect:
Non-uniform or Variable behavior only occurs strictly near the edges of the plates, or if the plates themselves are warped and non-parallel.
Force between two point charges is inversely proportional to the product of the charges and directly proportional to the square of the distance between them
B
Force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them
C
Force between two-point charges is directly proportional to the sum of the charges and inversely proportional to the square of the distance between them
D
Force between two-point charges is directly proportional to the product of the charges and the square of the distance between them
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
Coulomb's Law defines two distinct relationships: a numerator driving force magnitude up (charge), and a denominator driving it down exponentially (distance).
2. Formula:
$$ F = k \frac{q_1 q_2}{r^2} $$
3. Solution:
The formula states force is "directly proportional" to the numerator (the mathematical product of \( q_1 \) and \( q_2 \)).
The formula states force is "inversely proportional" to the denominator (the spatial square of the distance \( r^2 \)).
Statement B maps perfectly to this mathematical reality.
4. Why other options are incorrect:
Option A flips the numerator and denominator relationships. Option C falsely claims the charges are added (sum) rather than multiplied (product). Option D claims force scales up with distance, which is entirely false.
In an electric dipole composed of equal and opposite charges, the midpoint behaves uniquely regarding scalar (potential) and vector (field) quantities.
Potential is scalar. Since the distance to the positive charge equals the distance to the negative charge, their potentials sum algebraically to exactly zero: \( V_{net} = \frac{kq}{r} - \frac{kq}{r} = 0 \).
Electric field is a vector. The positive charge pushes a positive test charge right. The negative charge pulls that same test charge right. The vectors constructively add together, meaning the total field is non-zero (\( E \neq 0 \)).
4. Why other options are incorrect:
Option B assumes the fields cancel out (which only happens between two identical like-charges). Options C and D assume the scalar potential fails to cancel.
Q.201 A particle have a charge of 1C falls through a potential difference of 5 V. The energy required by it is:
A
\( 1.6 \times 10^{-19} \text{ J} \)
B
\( 5 \times 10^{-19} \text{ J} \)
C
6 J
D
5 J
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
The energy gained or work done on a massive charge accelerating through an electrostatic potential is the basic product of the charge and voltage.
2. Formula:
$$ E = q \Delta V $$
3. Solution:
Identify total charge: \( q = 1 \text{ C} \).
Identify voltage: \( V = 5 \text{ V} \).
Substitute and multiply natively: \( E = 1 \times 5 = 5 \text{ Joules} \).
4. Why other options are incorrect:
Options A and B attempt to trick the student into using the charge of a single electron (\( 1.6 \times 10^{-19} \text{ C} \)) instead of the explicitly stated 1 Coulomb macro-charge. Option C adds the values rather than multiplying.
Q.202 If the value of electric field intensity between the plates increases two times then energy stored in a capacitor becomes:
A
Double
B
Half
C
One Forth
D
Quadruple
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
The total electrical energy stored entirely within a region of space is dictated by its energy density, which scales with the square of the local electric field.
2. Formula:
$$ U = \frac{1}{2} \epsilon_0 E^2 (A \cdot d) $$
3. Solution:
The energy equation proves that potential energy (\( U \)) is directly proportional to the square of the electric field intensity (\( E^2 \)).
If the new field is \( 2E \), substituting it yields \( (2E)^2 = 4E^2 \).
The entire system's total stored energy increases by a factor of 4, quadrupling.
4. Why other options are incorrect:
Option A assumes a false linear relationship (\( U \propto E \)). Options B and C describe mathematically inverse relationships that apply to distance, not field strength.
Q.203 A fully charged capacitor with charge q and capacitance C is connected across a resistor R in series. Power dissipated across resistor will be zero after time:
A
\( t = 0 \)
B
\( t = RC \)
C
\( t = 5RC \)
D
\( t = \infty \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
During the discharge of a capacitor through a resistor, the outgoing current drops exponentially over time. Power dissipation requires current.
2. Formula:
$$ P = I^2 R \quad \text{where} \quad I(t) = I_0 e^{-t/RC} $$
3. Solution:
Because current decays according to the asymptotic function \( e^{-t/RC} \), the current only reaches exactly zero mathematically when \( t \to \infty \).
Since Power strictly relies on Current (\( P = I^2 R \)), power dissipation only truly halts at mathematical infinity, despite being practically negligible much earlier.
4. Why other options are incorrect:
At \( t = 0 \), power dissipation is at its absolute maximum. At 1 RC and 5 RC, current is still flowing (approx 37% and 1% of max current remain, respectively), meaning power is still being dissipated.
Q.204 A dielectric for a capacitor can be I. polar II. non-polar
A
II
B
both I and II
C
neither I nor II
D
I
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
Dielectric insulators are classified by the natural state of their molecules, but both types successfully polarize and function perfectly inside a capacitor.
Polar dielectrics (like water) have molecules with permanent dipole moments that physically rotate to align with the external field.
Non-polar dielectrics (like air or rubber) have symmetrical molecules, but the external field distorts their electron clouds, inducing temporary dipoles.
Both categories successfully generate an opposing internal electric field, fulfilling their role as dielectrics.
4. Why other options are incorrect:
Selecting only I or only II fails to recognize that the fundamental requirement of a dielectric is simply that it is an insulator capable of polarization, regardless of its baseline molecular symmetry.
Q.205 The electric potential at a point in an electric field is the amount of work done to move:
A
any amount of charge from infinity to that point
B
a unit positive charge from infinity to that point
C
any amount of charge from any position to that point
D
a unit negative charge from infinity to that point
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
This is the rigid, standardized textbook definition of absolute electric potential, creating a universal baseline for measurement.
2. Formula:
$$ V = \frac{W_{\infty \to r}}{q_0} $$
3. Solution:
To define "absolute" potential, the starting location must have precisely zero potential, which only occurs at an infinite spatial distance (\( r = \infty \)).
By long-standing physics convention, we normalize this work by standardizing it against a "unit positive test charge" (\( +1 \text{ C} \)).
4. Why other options are incorrect:
Moving 'any amount' of charge defines raw Work, not normalized Potential. Using 'any position' defines Potential Difference (\( \Delta V \)), not absolute potential. Standard convention prohibits using a negative test charge.
Q.206 A spherical liquid drop has a diameter of 2cm and is given a charge of 1mC. The potential at the surface of the drop is
A
900 MV
B
0.9 MV
C
0.45 MV
D
4.5 MV
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
The electric potential at the direct surface of a charged conducting sphere behaves identically to a point charge mathematically located at the sphere's absolute center.
Options C and D mistakenly utilize the full 2cm diameter instead of halving it for the radius. Option B suffers from a massive scientific notation decimal error.
Q.147 A capacitor of capacitance 'C' has a charge 'Q' and stored energy is 'W'. If the charge is increased to '2Q'. The stored energy will be:
A
2W
B
4W
C
W/4
D
W/2
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
The energy within a capacitor scales quadratically with the total amount of charge it possesses (provided its physical capacitance remains unchanged).
2. Formula:
$$ W = \frac{Q^2}{2C} $$
3. Solution:
The original stored energy equation is \( W = \frac{Q^2}{2C} \).
When charge is doubled to \( 2Q \), plug it in: \( W' = \frac{(2Q)^2}{2C} = \frac{4Q^2}{2C} \).
Factoring out the original expression yields \( W' = 4 \times W \).
4. Why other options are incorrect:
Option A falsely assumes energy scales linearly with charge (\( U \propto Q \)). Options C and D represent decreases, which physically contradicts forcing more charge into the plates.
Q.148 To store the electric charge the ultra-capacitors, use ____ effect.
A
Single layer
B
Double layer
C
Triple layer
D
Quadruple layer
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
Ultracapacitors (or supercapacitors) bridge the gap between traditional dielectric capacitors and batteries by utilizing electrochemical phenomena to achieve massive capacitances.
2. Formula:
$$ C = \frac{\epsilon A}{d} $$
3. Solution:
They achieve immense capacitance by utilizing the 'Electrochemical Double Layer' effect.
Instead of a solid dielectric, they use a porous electrolyte where charged ions gather incredibly close to the electrodes (down to nanometer distances, drastically shrinking \( d \) in the formula).
4. Why other options are incorrect:
Single, Triple, and Quadruple layers are scientifically fabricated distractors. The term of art in chemistry and physics for this specific boundary phenomenon is strictly 'Double Layer'.
Q.149 What is the potential difference between two points in an electric field if it takes 600J of energy of move a charge of 2 C between these two points?
A
1200 V
B
800 V
C
300 V
D
0 V
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
Potential difference (Voltage) directly measures the energetic 'cost' of forcing a set amount of charge from one location to another.
2. Formula:
$$ V = \frac{W}{q} $$
3. Solution:
Given the total Work (Energy): \( W = 600 \text{ J} \).
Given the total Charge: \( q = 2 \text{ C} \).
Substitute and evaluate: \( V = \frac{600}{2} = 300 \text{ V} \).
4. Why other options are incorrect:
Option A involves mathematically multiplying the values instead of dividing. Option D implies moving along an equipotential surface where work is zero.
Q.150 A charged particle moves in a uniform electric field between two oppositely charged parallel metal plates. To calculate the force acting on the particle due to the electric field, which quantity is not required?
A
Particle charge
B
Particle speed
C
Plate separation
D
Potential difference between the plates
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
The electrostatic force exerted on a charged particle inside a uniform electric field depends exclusively on the field's geometric parameters and the particle's own charge.
2. Formula:
$$ F = q E = q \left(\frac{V}{d}\right) $$
3. Solution:
Looking at the combined formula, computing Force requires: Particle charge (\( q \)), Potential difference (\( V \)), and Plate separation (\( d \)).
The particle's dynamic speed is entirely absent from the equations of electrostatics.
4. Why other options are incorrect:
Options A, C, and D are explicit variables within the \( F = q(V/d) \) equation, rendering them absolutely necessary for calculation. Speed only matters for magnetic fields (\( F = qvB \)).
The law treats the permittivity constant (\( \epsilon_0 \)) as a fundamental prerequisite input, an established property of the universe/medium.
You manipulate the formula to solve for the field (\( E \)), the flux (\( EA \)), or the enclosed charge (\( Q \)). You cannot "find" permittivity via this macroscopic law.
4. Why other options are incorrect:
The other options are variables algebraically locked into the primary equation, making them all directly solvable if the other parameters are known.
Q.152 Electric field intensity inside a hollow charged sphere is:
A
Zero
B
Maximum
C
Negative
D
Positive
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
In electrostatic equilibrium, any net charge placed upon a hollow conducting sphere migrates entirely to its outer exterior surface due to mutual repulsion.
If we define a Gaussian surface completely inside the hollow cavity of the sphere, the enclosed charge \( Q_{enc} \) is exactly zero.
Since \( Q_{enc} = 0 \), the integral mathematically dictates that the internal Electric field \( E \) must also be exactly zero everywhere inside.
4. Why other options are incorrect:
If the field were maximum, positive, or negative, loose charges inside the metal walls would be continually accelerated, breaking the definition of electrostatic equilibrium.
Q.153 As per Coulomb's law, the force of attraction or repulsion between two-point charges is directly proportional to the:
A
Sum of the magnitude of charges
B
Square of the distance between them
C
Product of the magnitude of charges
D
Cube of the distance
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
Coulomb derived mathematically that the strength of the electrostatic pull or push depends on the multiplied interaction of the two charge magnitudes involved.
2. Formula:
$$ F \propto q_1 \times q_2 $$
3. Solution:
The numerator of the Coulomb equation strictly dictates a multiplicative product (\( q_1 \cdot q_2 \)), meaning if you double one charge, the total mutual force doubles for both.
4. Why other options are incorrect:
Option A (sum) is algebraically incorrect. Options B and D refer to the spatial relationship, but force is inversely proportional to the square of the distance, not directly.
A perfect vacuum is exactly defined as \( \epsilon_r = 1.0000 \).
Standard dry air contains sparse gas molecules that polarize very slightly, raising its constant just barely above a vacuum to approximately \( 1.0006 \).
4. Why other options are incorrect:
Option A is the strict mathematical value for a pure vacuum. Options C and D represent denser liquid or solid dielectrics, not near-vacuum gases.
Q.155 Equivalent capacitance is less than least value of capacitance in a combination in:
A
Series
B
Parallel
C
Closed circuit
D
Open circuit
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
The mathematical method for summing capacitors in series relies on adding their reciprocals, which guarantees a final value strictly smaller than the smallest individual component.
Because you are adding fractions to form a larger denominator value, inverting that final sum forces the equivalent capacitance (\( C_{eq} \)) to shrink below any of the starting values. This effectively widens the total plate separation.
4. Why other options are incorrect:
In Parallel combinations (Option B), the capacitances are directly added (\( C_{eq} = C_1 + C_2 \)), so the equivalent value is always much larger than the largest individual capacitor.
Q.157 When dielectric is placed between the plates of a capacitor then capacitance will:
A
Decrease
B
Increase
C
Becomes half
D
Remains same
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
Inserting a dielectric medium allows the capacitor plates to hold more charge at the exact same voltage, fundamentally boosting its storage capacity.
2. Formula:
$$ C_{med} = \epsilon_r C_{vac} $$
3. Solution:
Because the dielectric constant (\( \epsilon_r \)) of any insulating material is always greater than 1 (vacuum), multiplying the baseline capacitance by \( \epsilon_r \) guarantees that the final capacitance mathematically increases.
4. Why other options are incorrect:
Decreasing the capacitance (Option A) would occur if you removed the dielectric. The other options contradict standard electrostatic enhancement.
Q.158 A one micro farad capacitor of a TV is subjected to 4000 V potential difference. The energy stored in capacitor is:
A
16
B
\( 4 \times 10^{-3} \text{ J} \)
C
\( 2 \times 10^{-3} \text{ J} \)
D
8 J
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
High voltages applied across standard capacitors result in significant electrical potential energy accumulation, proportional to the square of the voltage.
The positive and negative exponents cancel perfectly (\( 10^{-6} \times 10^6 = 1 \)).
Calculate final value: \( U = \frac{1}{2} \times 16 = 8 \text{ Joules} \).
4. Why other options are incorrect:
Option A forgets the \( 1/2 \) coefficient. Options B and C arise from failure to square the voltage correctly or from mishandling the scientific notation prefixes.
Q.159 A particle carrying a charge 2e falls through a P.D of 3 V. The energy acquired will be:
A
6 eV
B
7 eV
C
1.5 eV
D
5 eV
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
The electron-volt (eV) unit system bypasses the need for massive Coulombs conversions. The energy in eV is simply the numerical number of elementary charges multiplied by the potential difference.
2. Formula:
$$ E (\text{in eV}) = q (\text{in e}) \times V $$
3. Solution:
Charge of the particle: \( q = 2e \) (it has two elementary charges, like an alpha particle or divalent ion).
Voltage drop: \( V = 3 \text{ V} \).
Multiply them natively: \( E = 2e \times 3 \text{ V} = 6 \text{ eV} \).
4. Why other options are incorrect:
Option D represents adding the numbers (3+2). Option C represents dividing the numbers (3/2). Both represent fundamental algebraic misunderstandings of the energy relationship.
Q.160 Due to polarization process of dielectric, the electric intensity:
A
Increases
B
Decreases
C
Remains same
D
Becomes zero
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
When placed in an external electric field, the atoms inside a dielectric shift. Positive nuclei pull slightly one way, electron clouds the other. This creates a weak 'internal' field pointing backward, fighting the main field.
2. Formula:
$$ E_{net} = E_{external} - E_{internal} $$
3. Solution:
Because the newly created internal polarization field acts in direct opposition to the primary external field, their vector sum forces the overall net electric intensity inside the material to actively decrease.
4. Why other options are incorrect:
It does not become fully zero (Option D) unless the material is a perfect conductor that shifts free electrons until total cancellation occurs. Dielectrics only partially cancel the field.
Q.161 Two charges of magnitude \( q_1 = 1\mu\text{C} \) and \( q_2 = 5\mu\text{C} \), are separated at a distance \( r = 1 \times 10^{-3} \text{ m} \) apart, the ratio of the magnitude of the forces acting on them will be:
A
1:5
B
1:25
C
1:3
D
1:1
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
Newton's Third Law of Motion strictly dictates that all forces are mutual interactions. An electrostatic force is a shared bond between two objects.
2. Formula:
$$ F_1 = F_2 = k \frac{q_1 q_2}{r^2} $$
3. Solution:
Despite \( q_2 \) being five times larger in magnitude than \( q_1 \), the actual force formula multiplies them together.
The force \( q_1 \) feels is \( k(1)(5)/r^2 \).
The force \( q_2 \) feels is \( k(5)(1)/r^2 \).
Both forces are mathematically identical, making their ratio perfectly 1:1.
4. Why other options are incorrect:
Option A attempts to bait students into comparing the ratio of the charges themselves, rather than the mutual forces they experience.
Q.162 Two equal and opposite charges of 10C are separated at a distance of 10 cm, the electric potential at mid-point between the charges is
A
20 V
B
10 V
C
5 V
D
0 V
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
Electric potential is a scalar quantity, not a vector. To find total potential at a point, you simply add the algebraic values of the potentials from all nearby charges.
2. Formula:
$$ V_{net} = V_+ + V_- $$
3. Solution:
At the exact midpoint, the distance to the positive charge is \( r \) and the distance to the negative charge is an identical \( r \).
The positive charge creates a potential of \( V_+ = + \frac{kq}{r} \).
The negative charge creates a potential of \( V_- = - \frac{kq}{r} \).
Because they are identical in magnitude but opposite in sign, their algebraic sum evaluates to precisely \( 0 \text{ V} \).
4. Why other options are incorrect:
If this question asked for Electric Field (a vector), the fields would reinforce and add together. But for Potential, opposite signs flawlessly cancel out.
Q.163 The force of repulsion between two alike charges is 10N in vacuum. When a material of \( \epsilon_r = 2 \) is placed between them, new force will be:
A
20 N
B
15 N
C
10 N
D
5 N
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
Inserting a dielectric material between two free charges shields them from each other, dampening the electrostatic interaction force between them.
2. Formula:
$$ F_{medium} = \frac{F_{vacuum}}{\epsilon_r} $$
3. Solution:
Original force in vacuum: \( F_{vac} = 10 \text{ N} \).
Dielectric constant of new medium: \( \epsilon_r = 2 \).
The y-axis represents Charge (\( Q \)), and the x-axis represents Time (\( t \)).
The mathematical rate of flow of charge over time (\( \Delta Q / \Delta t \)) is the fundamental definition of Electric Current (\( I \)).
4. Why other options are incorrect:
Voltage relates charge to capacitance (\( V = Q/C \)). Energy relates charge to voltage (\( U = \frac{1}{2} QV \)). Only current involves a time derivative.
Q.165 Two small charged objects attract each other with a force F when separated by a distance d. If the charge on each object is reduced to q/2 the force becomes:
A
F/16
B
F/8
C
F/4
D
F/2
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
The electrostatic force relies on the multiplied product of both interacting charges. Changing either charge individually affects the final total force.
2. Formula:
$$ F = k \frac{q_1 q_2}{d^2} $$
3. Solution:
Original force: \( F = k(q)(q)/d^2 \).
Both charges are individually halved, becoming \( q/2 \).
Substitute new charges: \( F' = k \frac{(q/2)(q/2)}{d^2} \).
Pull out the fractions: \( F' = \frac{1}{4} \left( k \frac{q_1 q_2}{d^2} \right) \).
The resulting force is exactly \( F/4 \).
4. Why other options are incorrect:
Option D occurs if only one of the charges is halved. Option A would occur if the distance was also accidentally doubled alongside the charge reductions.
Q.168 The change in electric potential with respect to distance equals to:
A
Potential gradient
B
Amount of the charge
C
Potential difference
D
Surface charge density
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
In physics, any quantity's mathematical "rate of change with respect to distance" is referred to as its spatial gradient.
2. Formula:
$$ E = - \frac{\Delta V}{\Delta r} $$
3. Solution:
The term \( \Delta V / \Delta r \) evaluates how steeply the voltage drops off as you move through space.
This specific mathematical construct is formally titled the "Potential Gradient," and it is strictly equivalent to the magnitude of the Electric Field.
4. Why other options are incorrect:
Potential difference is just \( \Delta V \) without dividing by distance. Surface charge density is charge per unit area (\( Q/A \)).
Q.169 The amount of energy required in moving an electron of charge (e) by the application of 1volt potential difference equals to
A
1 keV
B
1 MeV
C
1 GeV
D
1 eV
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
This question is literally stating the textbook definition of the electron-volt unit, a standard energy metric used heavily in quantum mechanics and particle physics.
2. Formula:
$$ E = q \Delta V $$
3. Solution:
When \( q \) equals exactly 1 fundamental electron charge (\( 1e \)).
And \( \Delta V \) equals exactly 1 Volt.
The mathematical result is \( 1e \times 1\text{V} = 1 \text{ eV} \) (One electron-volt).
4. Why other options are incorrect:
The other options include metric prefixes: kilo (1000 eV), Mega (1,000,000 eV), and Giga (1,000,000,000 eV), which apply to much higher energy particle accelerators, not a simple 1-volt drop.
Q.170 Which law explains the relation between amount of charges and force between them?
A
Ohm's law
B
Lenz's law
C
Coulomb's law
D
Ampere's law
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
Charles-Augustin de Coulomb formulated the foundational electrostatic law that quantifies the exact force between two stationary, electrically charged particles.
2. Formula:
$$ F = k \frac{q_1 q_2}{r^2} $$
3. Solution:
Coulomb's equation directly bridges the gap between measuring static charge magnitude and measuring dynamic mechanical force.
4. Why other options are incorrect:
Ohm's law bridges Voltage and Current in circuits. Lenz's law dictates the direction of induced magnetic currents. Ampere's law equates magnetic fields to wire currents.
Q.171 What does the electric field around a charge represent?
A
Size of the charge
B
Effective area for electrostatic force
C
Path followed by the charge
D
Speed of the charge
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
Historically, an electric field is described as an invisible 'sphere of influence' permeating the space surrounding a source charge.
2. Formula:
$$ \vec{F} = q_0 \vec{E} $$
3. Solution:
The field dictates the strict geographic "effective area" (or more accurately, the 3D volume) within which another charged particle will successfully "feel" an electrostatic pushing or pulling force.
4. Why other options are incorrect:
The field does not represent physical size (Option A). Field lines can imply a path, but the field itself represents the force capability, not the motion (Option C). Electrostatics explicitly ignores speed (Option D).
The standard unit of Electric Field Intensity can be mathematically proven via two separate defining equations (Force vs Gradient) to be entirely interchangeable.
2. Formula:
$$ E = \frac{F}{q} \implies \text{N/C} \quad \text{and} \quad E = \frac{V}{d} \implies \text{V/m} $$
3. Solution:
Start with Volts per meter (V/m).
Expand Volt into Joules per Coulomb (J/C): \( \frac{\text{J/C}}{\text{m}} = \frac{\text{J}}{\text{C} \cdot \text{m}} \).
Q.173 Which statement is the best for electric flux of a point charge?
A
Independent of charge
B
Independent of medium
C
Independent of shape of surface in which charge is enclosed
D
Independent position of charge in space
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
According to Gauss's Law, the total outward electrical flux piercing any closed imaginary surface relies only on the amount of charge trapped inside, regardless of how chaotic or asymmetrical the boundary's physical geometry is.
2. Formula:
$$ \Phi_E = \frac{Q_{enc}}{\epsilon_0} $$
3. Solution:
Because the integral of a closed surface fundamentally sums all outward vector lines, reshaping the boundary just changes where the lines exit, not the total count of exiting lines. Thus, flux is completely independent of the enclosing shape.
4. Why other options are incorrect:
Flux directly depends on the magnitude of the charge (ruling out A). It depends heavily on the medium's permittivity \( \epsilon_0 \) (ruling out B).
Q.174 Two capacitors \( 6 \mu\text{F} \) and \( 12 \mu\text{F} \) are in series connected across a 200 volts D.C supply calculate the charges on each capacitor, respectively.
For any circuit components strung in series, the identical total current must flow through them, guaranteeing that series capacitors hold the exact same magnitude of charge regardless of their individual capacities.
Adjusting the decimal for scientific notation yields \( 8 \times 10^{-4} \text{ C} \).
Since they are in series, BOTH capacitors possess this identical exact charge.
4. Why other options are incorrect:
Options B and C falsely assume capacitors in series possess different charges. Option D makes a catastrophic error regarding the 'micro' prefix, scaling the answer by a million.
Q.175 While studying charging and discharging of a capacitor, RC = Resistance \( \times \) capacitance is known as?
A
Electrostatic constant
B
Time constant
C
Dielectric constant
D
Proportionality constant
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
In any RC circuit, the mathematical product of the Resistor's resistance and the Capacitor's capacitance specifically dictates how rapidly the circuit responds to changes in voltage.
2. Formula:
$$ \tau = R \times C $$
3. Solution:
Because Ohm \( \times \) Farad perfectly yields Seconds, this product provides a physical metric for time.
This specific value (\( \tau \)) is formally titled the "Time Constant," marking the exact time required to charge to ~63% capacity.
4. Why other options are incorrect:
Dielectric constant (\( \epsilon_r \)) is an insulation property. Electrostatic constant (\( k \)) is used in Coulomb's Law.
Q.176 In parallel combination of two capacitors, their equivalent capacitance is equal to:
A
\( C_1 + C_2 \)
B
\( 1/C_1 + 1/C_2 \)
C
\( \frac{C_1 C_2}{C_1 + C_2} \)
D
\( \frac{2C_1 C_2}{C_1 + C_2} \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
Wiring capacitors in parallel inherently fuses their isolated metal plates into one massive, shared conductive area, which directly aggregates their total individual storage capacities.
2. Formula:
$$ C_{eq} = C_1 + C_2 + C_3 + \dots $$
3. Solution:
Because parallel branches each feel the same total battery voltage, the total combined charge stored by the system is simply \( Q_{total} = Q_1 + Q_2 \).
Substituting \( Q = CV \) proves that \( C_{eq}V = C_1V + C_2V \), allowing the shared voltage to cancel out, leaving straightforward addition.
4. Why other options are incorrect:
Options B and C represent the formulas applied specifically to Capacitors wired strictly in Series, where capacity acts reciprocally.
Q.177 The SI unit of capacitance of a capacitor is:
A
Coulomb
B
Volt
C
Farad
D
Ampere
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
The standard unit of electrical capacitance is named in honor of the English physicist Michael Faraday, honoring his contributions to electromagnetism.
Q.121 A capacitor is charged with a battery and energy stored is U. After disconnecting battery another capacitor of same capacity is connected in parallel to the first capacitor. Then energy stored in each capacitor is
A
U/2
B
U/4
C
4U
D
2U
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
When a charged capacitor is disconnected, its total charge \( Q \) becomes trapped and strictly conserved. Connecting an identical uncharged capacitor in parallel forces the charge to distribute evenly.
2. Formula:
$$ U = \frac{Q^2}{2C} $$
3. Solution:
Initial state: Single capacitor has charge \( Q \). Energy is \( U = \frac{Q^2}{2C} \).
Second state: Because they are identical and parallel, the trapped total charge \( Q \) splits perfectly. The new charge on EACH individual capacitor is \( Q' = Q/2 \).
Calculate the new energy for ONE capacitor: \( U_{each} = \frac{(Q/2)^2}{2C} = \frac{Q^2/4}{2C} = \frac{1}{4} \left( \frac{Q^2}{2C} \right) \).
Since the original \( U = \frac{Q^2}{2C} \), the final energy in each is exactly \( U/4 \).
4. Why other options are incorrect:
Option A (U/2) is the total combined energy left in the entire system, representing the 50% energy lost to heat during charge redistribution. The question asks for the energy in each capacitor, which is half of the remaining total.
Q.122 What is the potential difference between two points in an electric field if it takes 600J of energy to move a charge of 2C between these two points?
A
1200 V
B
800 V
C
300 V
D
0 V
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
Potential difference is a direct measure of work accomplished or required per unit of charge.
2. Formula:
$$ V = \frac{W}{q} $$
3. Solution:
Given work energy: \( W = 600 \text{ J} \).
Given charge: \( q = 2 \text{ C} \).
Substitute and divide: \( V = \frac{600}{2} = 300 \text{ V} \).
4. Why other options are incorrect:
Option A results from incorrectly multiplying 600 by 2 instead of dividing.
The law utilizes electric flux (\( \Phi \)), charge (\( Q \)), and field intensity (\( E \)) as its primary mathematical variables.
Permittivity (\( \epsilon_0 \)) is a universal physical constant (or a fixed medium property). It is a given prerequisite parameter required to use the law, rather than an unknown physical quantity you utilize the law to solve for.
4. Why other options are incorrect:
By rearranging the Gauss's Law equation, one can easily isolate and solve for field intensity, flux, or enclosed charge.
Q.124 Which one of the following statements is true?
A
Electrostatic force obeys inverse square law while gravitational force does not
B
Both gravitation force and electrostatic force are repulsive in nature
C
Gravitational force is much weaker than electrostatic force
D
Both electrostatic force and gravitational force don't obey inverse square law
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
Comparing the fundamental forces of nature: Electrostatic forces are immensely powerful compared to Gravity on the scale of fundamental particles.
2. Formula:
$$ F_E = k \frac{q_1 q_2}{r^2} \quad \text{vs} \quad F_g = G \frac{m_1 m_2}{r^2} $$
3. Solution:
Statement A is false: Both forces mathematically feature a \( 1/r^2 \) drop-off, thus both strictly obey the inverse square law.
Statement B is false: Gravity only attracts; it has no physically proven repulsive counterpart.
Statement C is true: The electrostatic force between an electron and proton is roughly \( 10^{39} \) times stronger than their mutual gravitational pull. Gravity is the weakest fundamental force.
4. Why other options are incorrect:
They contradict the fundamental mathematical structure and empirical strength hierarchies of Newton's and Coulomb's laws.
Coulomb's constant \( k \) is not a true universal constant like \( c \) (speed of light). Its numerical output relies entirely on the intervening environment and how humans choose to measure it.
2. Formula:
$$ k = \frac{1}{4 \pi \epsilon} $$
3. Solution:
The permittivity term (\( \epsilon \)) is dictated by the physical nature of the medium (vacuum vs water vs air).
The value of \( k \) changes if measured in SI units (\( 9 \times 10^9 \)) versus CGS units (where \( k \) equals exactly 1). Thus, it heavily depends on the system of units.
4. Why other options are incorrect:
A and B are incomplete on their own. Option C is false because the constant behaves identically regardless of whether the charges are positive or negative.
Q.126 A charged particle is moving in a uniform electric field. For the motion of the particle due to the field, which quantity has a constant non-zero value?
A
Acceleration
B
Displacement
C
Rate of change of acceleration
D
Velocity
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
A uniform electric field provides an unchanging, constant electrostatic force on a charged particle at all points in space.
Because both charge (\( q \)) and field strength (\( E \)) are constant, the resulting force (\( F \)) is totally constant.
By Newton's Second Law, a constant net force results in a constant, unchanging acceleration vector.
4. Why other options are incorrect:
Because the particle is accelerating, its velocity is continuously changing (ruling out D). Displacement is also rapidly changing as it moves (ruling out B). The rate of change of acceleration (jerk) is precisely zero, not non-zero (ruling out C).
Q.127 A capacitor of capacitance 'C' has a charge 'Q' and stored energy is 'W'. If the charge is increased to '2Q'. The stored energy will be:
A
2W
B
4W
C
W/4
D
W/2
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
The electrical potential energy stored in a capacitor is directly proportional to the square of the charge it holds, assuming capacitance remains constant.
2. Formula:
$$ W = \frac{Q^2}{2C} $$
3. Solution:
Let the initial energy be \( W = \frac{Q^2}{2C} \).
Substitute the new charge \( 2Q \) into the equation: \( W' = \frac{(2Q)^2}{2C} \).
Q.128 Electric intensity due to charge distributions are calculated using which following law?
A
Ohm's law
B
Faraday's law
C
Gauss's law
D
Ampere's law
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
Gauss's law relates the electric flux passing through a closed surface to the charge enclosed by that surface, providing a powerful tool to calculate electric fields (intensity) for symmetrical charge distributions.
By identifying a symmetrical Gaussian surface, the integral simplifies, allowing for the direct algebraic isolation and calculation of the electric field intensity \( E \).
4. Why other options are incorrect:
Ohm's law relates voltage, current, and resistance. Faraday's law deals with electromagnetic induction. Ampere's law calculates magnetic fields from electric currents.
Q.129 The capacitance of capacitor does not depend on:
A
Area of plates
B
Medium
C
Distance between plates
D
Thickness of plates
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
The physical capacity of a parallel plate capacitor to store charge is dictated purely by the geometry of the facing surfaces and the insulating material between them.
2. Formula:
$$ C = \frac{\epsilon A}{d} $$
3. Solution:
The formula explicitly relies on Area (\( A \)), Separation Distance (\( d \)), and the Permittivity of the Medium (\( \epsilon \)).
The thickness of the conducting metal plates themselves is mathematically irrelevant to the electrostatic field established in the gap.
4. Why other options are incorrect:
Changing the area, distance, or dielectric medium will physically alter the capacitance value, making options A, B, and C dependent variables.
Electric potential describes the amount of potential energy per unit of charge at a specific point in space, which equates to the work required to bring that charge from infinity.
2. Formula:
$$ V = \frac{W}{q} $$
3. Solution:
The formula strictly states that Potential (\( V \)) is the Work (\( W \)) accomplished divided by the Charge (\( q \)).
4. Why other options are incorrect:
Force per unit charge defines the Electric Field Intensity (\( E \)). Power per unit charge is not a standard electrostatic quantity.
An electron-volt is the amount of kinetic energy gained by a single electron accelerating from rest through an electric potential difference of exactly one volt.
2. Formula:
$$ E = q \Delta V $$
3. Solution:
The charge of a single electron is \( q = 1.602 \times 10^{-19} \text{ C} \).
Multiply by 1 Volt: \( 1 \text{ eV} = (1.602 \times 10^{-19} \text{ C}) \times (1 \text{ V}) \).
Since \( 1 \text{ C} \cdot 1 \text{ V} = 1 \text{ J} \), the result is \( 1.602 \times 10^{-19} \text{ J} \).
4. Why other options are incorrect:
The other options feature incorrectly shifted decimal places. The fundamental charge of an electron strictly dictates the \( 1.602 \) coefficient.
Q.134 Two charges one of which is \( Q_1 = 3\mu\text{C} \) and second one is \( Q_2 = -1\mu\text{C} \) are separated by 100 cm. The electric potential is zero at a point
A
25 cm from \( Q_2 \)
B
75 cm from \( Q_2 \)
C
50 cm from \( Q_1 \)
D
33.3 cm from \( Q_1 \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
Electric potential is a scalar quantity. The total potential is zero where the positive potential from \( Q_1 \) exactly cancels the negative potential from \( Q_2 \).
Q.135 The electric field intensity at a point situated 4 m from a point charge is 200 N/C. If the distance is reduced to 2 m, what will be field intensity?
A
400 N/C
B
800 N/C
C
600 N/C
D
1200 N/C
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
The electric field of a point charge obeys the inverse square law. Halving the distance from the source dramatically magnifies the field strength.
2. Formula:
$$ E \propto \frac{1}{r^2} $$
3. Solution:
Original state: \( E_1 = 200 \text{ N/C} \) at \( r_1 = 4 \text{ m} \).
New state: \( r_2 = 2 \text{ m} \), which means the distance is halved (\( r_2 = r_1 / 2 \)).
Because \( E \) varies inversely with the square of the distance, replacing \( r \) with \( r/2 \) introduces a factor of \( 1 / (1/2)^2 = 1 / (1/4) = 4 \).
The new field is 4 times stronger: \( 200 \times 4 = 800 \text{ N/C} \).
4. Why other options are incorrect:
Option A incorrectly assumes a linear relationship (distance halved = field doubled). Options C and D are mathematically unrelated to the inverse-square geometric scaling.
Q.136 Which type of energy does the capacitor store?
A
Kinetic
B
Potential
C
Thermal
D
Light
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
A capacitor functions by separating charges, doing work against the electric field, and holding those separated charges static until a circuit is completed.
2. Formula:
$$ U_E = \frac{1}{2} C V^2 $$
3. Solution:
Because the charges are held stationary (static) on the opposing plates, the energy is strictly stored as Electrical Potential Energy inside the electric field.
4. Why other options are incorrect:
Kinetic energy involves active motion (like current flowing). Thermal and Light energy describe forms of energy dissipation (like a resistor or bulb), not storage.
Q.138 In equation \( E = - \frac{\Delta V}{\Delta r} \). Negative sign indicates that:
A
'E' point towards increasing 'V'
B
'E' point towards decreasing 'V'
C
'E' is perpendicular 'V'
D
'E' is parallel to '\( \Delta V \)'
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
The electric field vector inherently points from regions of positive charge (high potential) toward regions of negative charge (low potential).
2. Formula:
$$ \vec{E} = - \nabla V $$
3. Solution:
The potential gradient formula equates the electric field to the spatial rate of change of voltage.
The negative sign serves purely as a directional indicator, proving that if you walk strictly in the direction of the Electric field (\( E \)), the voltage (\( V \)) will steadily decrease.
4. Why other options are incorrect:
If E pointed towards increasing V, the equation would be positive. Field lines and equipotential lines are perpendicular, but this equation describes the gradient line itself, not the equipotential surface.
Q.139 The electric field intensity between two uniformly, oppositely charged parallel plates is:
A
\( \frac{\sigma}{2\epsilon_o} \)
B
\( \frac{\epsilon_o}{\sigma} \)
C
\( \frac{2\sigma}{\epsilon_o} \)
D
\( \frac{\sigma}{\epsilon_o} \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
The total electric field between two opposite parallel plates is the additive sum of the uniform fields generated independently by each plate.
2. Formula:
$$ E = E_{positive} + E_{negative} $$
3. Solution:
A single infinite charged plate generates a field of \( E_1 = \frac{\sigma}{2\epsilon_0} \).
Between two oppositely charged plates, the field vectors point in the exact same direction, reinforcing each other.
Total Field: \( E = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{2\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_o} \).
4. Why other options are incorrect:
Option A is the field strength for only a single sheet of charge, representing the region outside the plates or if one plate were removed. Options B and C are dimensionally inverted or incorrectly scaled.
Q.140 If \( 2\mu\text{F} \) and \( 4\mu\text{F} \) capacitors are connected in series. Their equivalent capacitance will be:
A
\( 8 \mu\text{F} \)
B
\( 0.75 \mu\text{F} \)
C
\( 1.33 \mu\text{F} \)
D
\( 6 \mu\text{F} \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
When capacitors are linked in series, their combined effective plate distance effectively increases, which reduces the total capacitance. We use the reciprocal sum rule.
2. Formula:
$$ C_{eq} = \frac{C_1 \times C_2}{C_1 + C_2} $$
3. Solution:
Identify the values: \( C_1 = 2 \mu\text{F} \) and \( C_2 = 4 \mu\text{F} \).
Multiply them for the numerator: \( 2 \times 4 = 8 \).
Add them for the denominator: \( 2 + 4 = 6 \).
Divide: \( C_{eq} = 8 / 6 = 1.33 \mu\text{F} \).
4. Why other options are incorrect:
Option D (6) is the answer if they were connected in parallel (simply adding them). Option B (0.75) is the inverted result (6/8), common if a student forgets to flip the final reciprocal fraction.
Q.141 A proton of charge \( 1.6 \times 10^{-19} \text{ C} \) is moved through a P.D of 20 V. The energy gained by the proton is:
A
\( -32 \times 10^{-19} \text{ J} \)
B
\( 32 \times 10^{-19} \text{ J} \)
C
\( -32 \times 10^{19} \text{ J} \)
D
\( 32 \times 10^{19} \text{ J} \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
The kinetic energy acquired by a charge moving freely through an electrostatic potential difference is simply the mathematical product of the charge and the voltage.
A proton is positively charged, and 'gaining' energy implies a positive increase in kinetic energy, matching the positive value.
4. Why other options are incorrect:
Option A introduces a negative sign, which would apply if calculating the loss of potential energy or moving an electron against the field. Options C and D utilize catastrophically massive positive exponents.
Q.142 Two similar point charges, each of 1 coulomb, placed in free space 1m apart, repel each other with a force of:
A
\( 8.85 \times 10^{-12} \text{ N} \)
B
\( 1.602 \times 10^{-19} \text{ N} \)
C
1 N
D
\( 9 \times 10^9 \text{ N} \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
This setup represents the foundational definition of Coulomb's constant \( k \). Finding the force between two exactly 1 Coulomb charges separated by exactly 1 meter effectively outputs the constant itself.
Option A is the permittivity of free space (\( \epsilon_0 \)), not the force constant. Option B is the charge of a single electron. Option C naively assumes the variables multiply without the \( k \) constant multiplier.
Q.143 If a slab of dielectric constant \( \epsilon_r = 2 \) is inserted between the parallel plates of a charged \( 20 \mu\text{F} \) capacitor. Its capacitance will:
A
Remain the same
B
Be four times
C
Be doubled
D
Be halved
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
Introducing an insulating dielectric material between capacitor plates reduces the internal electric field via polarization, which fundamentally acts to multiply and increase the total capacitance.
2. Formula:
$$ C_{med} = \epsilon_r C_{vac} $$
3. Solution:
Original vacuum capacitance: \( C_{vac} = 20 \mu\text{F} \).
Dielectric constant of new slab: \( \epsilon_r = 2 \).
Because 40 is twice the original 20, the capacitance has precisely doubled.
4. Why other options are incorrect:
Option D occurs if you erroneously divide by the dielectric constant (which happens to field strength, not capacitance). Option B is a distractor for square-law calculations.
The electrical potential energy is calculated by integrating the voltage across the capacitor as charge progressively builds up from zero to its maximum.
2. Formula:
$$ dU = V \cdot dQ \implies \int_0^Q \frac{q}{C} dq $$
Substituting \( Q = CV \) into that result gives the standard formula: \( E = \frac{1}{2} C V^2 \).
4. Why other options are incorrect:
Option C is missing the integral's \( 1/2 \) coefficient, representing a capacitor charging instantly at full voltage rather than ramping up. Option D evaluates strictly to Charge (\( Q \)), not Energy.
Electric field intensity describes the local strength of an electric field, quantified by measuring how hard it pushes or pulls on an established test charge.
2. Formula:
$$ \vec{E} = \frac{\vec{F}}{q} $$
3. Solution:
By mapping exactly to the formula, the vector Field \( \vec{E} \) is equal to Force (Newtons) divided by Charge (Coulombs).
4. Why other options are incorrect:
Force per unit mass (Option A) defines the Gravitational Field strength (\( g \)). Tesla is a magnetic unit. Watt is a power unit.
A capacitor is specifically engineered to hold an electrostatic field between isolated parallel plates, fundamentally serving as a short-term energy reservoir.
2. Formula:
$$ U_E = \frac{1}{2} C V^2 $$
3. Solution:
When connected to a source, charges accumulate on the plates, establishing an electric field that locks in electrical potential energy until discharged.
4. Why other options are incorrect:
An Inductor stores energy in a Magnetic field, not an electric potential field. A Generator converts mechanical energy into electrical energy dynamically, rather than statically storing it.
Q.114 Electric force and electric field intensity for a charge are
A
Parallel to each other
B
Opposite to each other
C
Perpendicular to each other
D
At any orientation
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
The electric field is explicitly defined based on the force experienced by a positive test charge. By standard physics convention, unless a negative sign is explicitly provided, we assume standard positive charge behavior.
2. Formula:
$$ \vec{F} = q \vec{E} $$
3. Solution:
Because \( q \) is a scalar multiplier, the resulting vector \( \vec{F} \) must lie along the exact same line as \( \vec{E} \). Thus, they are inherently parallel.
4. Why other options are incorrect:
They are anti-parallel (opposite) strictly for negative charges, but 'parallel' represents the baseline mathematical alignment of the vectors.
Q.115 A charge Q placed at the centre of two charges + q and + q. the system is in equilibrium then net force experienced by Q.
A
Zero
B
F
C
At equilibrium, net force will be zero
D
F/2
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
If a charge is positioned perfectly midway between two identical source charges, the electric fields from the two sources cancel each other out completely.
Find a common denominator to add the fractions: \( \frac{1}{C_{eq}} = \frac{C_2 + C_1}{C_1 C_2} \).
To find the actual equivalent capacitance (\( C_{eq} \)), you must invert the entire fraction.
Result: \( C_{eq} = \frac{C_1 C_2}{C_1 + C_2} \) (Product over Sum rule).
4. Why other options are incorrect:
Option A is the formula for Parallel capacitors. Option C is mathematically \( 1/C_{eq} \), forgetting the final necessary inversion. Option D is an invalid algebraic mashup.
This is the fundamental macroscopic definition of voltage (electric potential).
2. Formula:
$$ V = \frac{W}{q} $$
3. Solution:
By defining Joules of work (W) required to move a specific Coulomb count (q), we arrive at Joules/Coulomb. This ratio defines the Volt, which measures Electric Potential.
4. Why other options are incorrect:
Electric field is force per unit charge (Newtons per Coulomb), entirely lacking the distance metric required for 'work'.
An equipotential surface is a region in space where every single geometric point has the exact same voltage.
2. Formula:
$$ W = q \Delta V $$
3. Solution:
Because the surface has identical voltage everywhere, the difference between any point A and point B is strictly zero (\( \Delta V = V_B - V_A = 0 \)).
Therefore, \( W = q \times 0 = 0 \). No net work is done moving across it.
4. Why other options are incorrect:
Positive or negative work requires moving a charge up or down a potential gradient, which does not exist along the surface.
The potential difference between the spheres is \( \Delta V = \frac{Q}{4 \pi \epsilon_0} \left(\frac{1}{a} - \frac{1}{b}\right) \).
Find a common denominator for the radius terms: \( \left(\frac{b - a}{ab}\right) \).
Inverting this when dividing \( Q \) by \( \Delta V \) shifts \( ab \) to the numerator, resulting in \( C = 4 \pi \epsilon_0 \left(\frac{ab}{b-a}\right) \).
4. Why other options are incorrect:
The other options fail to account for the reciprocal geometric derivation resulting from the point-charge potential formula.
The energy stored in a parallel plate capacitor is \( 24\text{ J} \). What is the potential difference across its plates if the capacitance is \( 3\mu\text{F} \)?
A
16 kV
B
54 kV
C
8 kV
D
4 kV
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Calculate potential difference from electrostatic potential energy stored in a capacitor's electric field.
Q.109 The values of electric intensity will ____ due to the presence of dielectric medium:
A
Increase
B
Increase exponentially
C
Decrease
D
Remains same
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
A dielectric material consists of atoms that polarize when placed in an electric field. This polarization establishes an internal opposing electric field, which acts to reduce the overall net electric intensity.
2. Formula:
$$ E_{medium} = \frac{E_{vacuum}}{\epsilon_r} $$
3. Solution:
Because the relative permittivity (\( \epsilon_r \)) of any real dielectric is strictly greater than 1, dividing the original vacuum field by \( \epsilon_r \) guarantees that the final field strength mathematically decreases.
4. Why other options are incorrect:
Increasing the field would violate conservation of energy and require the material to actively pump energy into the field, which passive insulators cannot do.
Q.110 In capacitors, energy is stored in the form of:
A
Gravitational energy
B
Kinetic energy
C
Electric intensity
D
Magnetic energy
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
Capacitors physically store electrical potential energy in the space directly between their conducting plates.
2. Formula:
$$ U = \frac{1}{2} \epsilon_0 E^2 (A \cdot d) $$
3. Solution:
The energy density equation proves that the energy fundamentally resides within the electric field (often referred to as 'electric intensity' in older nomenclature) that bridges the gap between the isolated charges.
4. Why other options are incorrect:
Magnetic energy is strictly stored by Inductors (via current flow). Kinetic energy pertains to motion. Gravitational energy is mass-dependent, not charge-dependent.
Q.112 By increasing area of the plates and decreasing distance between them the capacitance of capacitor:
A
Increases
B
Decreases
C
Remains unchanged
D
Depending upon temperature
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
Capacitance describes a plate's ability to hold charge. A larger surface area allows more charges to gather without repelling each other. A tighter distance increases the attractive pull between opposite plates, allowing even more charge to hold.
2. Formula:
$$ C = \frac{\epsilon_0 A}{d} $$
3. Solution:
If numerator Area (\( A \)) is increased, \( C \) increases.
If denominator distance (\( d \)) is decreased, \( C \) also independently increases.
Doing both simultaneously compounds the effect, resulting in a large guaranteed increase in total Capacitance.
4. Why other options are incorrect:
Decreasing occurs if you do the exact opposite (decrease area, widen gap). Temperature generally has trivial to no effect on physical plate geometry.
Q.105 Electric field strength of a point charge is E and electric potential is V at a distance r from the point charge. What is the electric potential at a point for the same point charge where electric field strength is E/4?
A
V/4
B
4V
C
V/2
D
2V
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
For a point charge, the electric field (\( E \)) decays by an inverse-square law, while the electric potential (\( V \)) decays inversely with linear distance.
2. Formula:
$$ E \propto \frac{1}{r^2} \quad \text{and} \quad V \propto \frac{1}{r} $$
3. Solution:
If the new field is \( E/4 \), this means the distance must have been doubled (since \( 1/(2r)^2 = 1/4r^2 \)). Let the new distance be \( r' = 2r \).
Since potential depends on \( 1/r \), applying the new doubled distance gives \( V' = \frac{k q}{2r} = \frac{1}{2} \left( \frac{k q}{r} \right) \).
Therefore, the new potential is \( V/2 \).
4. Why other options are incorrect:
Option A falsely assumes V decreases at the identical squared rate as E. Option D incorrectly implies that moving further away increases the potential.
Q.106 A particle carrying a charge of \( 5e \) falls through a potential difference of 25V. What would be energy acquired by the particle in 'J'.
A
\( 125 \times 10^{-19} \text{ J} \)
B
\( 1.6 \times 10^{-19} \text{ J} \)
C
\( 125 \times 1.6 \times 10^{-19} \text{ J} \)
D
125 J
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
The energy gained by a charge moving across a potential drop is equal to the product of charge and voltage, where fundamental unit 'e' must be expanded to its Coulomb value.
2. Formula:
$$ Energy = q \Delta V $$
3. Solution:
Identify the charge: \( q = 5e \).
Identify the voltage: \( V = 25 \text{ V} \).
Energy in electron-Volts (eV): \( E = (5e) (25 \text{ V}) = 125\text{ eV} \).
To convert eV to Joules, substitute \( e = 1.6 \times 10^{-19} \text{ C} \).
Energy in Joules: \( E = 125 \times 1.6 \times 10^{-19} \text{ J} \).
4. Why other options are incorrect:
Option A misses the 1.6 multiplier entirely. Option B is simply the charge of a single electron. Option D assumes macroscopic Coulombs rather than elementary charges.
Q.107 Electric field strength at a point between oppositely charge plates is E. If the distance between plates is reduced to half, what will be the new value of electric intensity?
A
4E
B
E/2
C
E/4
D
2E
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
For parallel plates held at a constant potential difference, the uniform electric field is directly proportional to voltage and inversely proportional to plate separation.
2. Formula:
$$ E = \frac{V}{d} $$
3. Solution:
Let the original distance be \( d \) and the original field be \( E \).
The new distance is \( d' = d / 2 \).
Substitute into the formula: \( E_{new} = \frac{V}{d/2} = 2 \left(\frac{V}{d}\right) = 2E \).
4. Why other options are incorrect:
Option B results from a direct proportional misinterpretation (thinking decreasing distance decreases the field). Options A and C confuse uniform field principles with point-charge inverse-square laws.
Q.97 Coulomb's law is given by the formula \( F=k \frac{q_1 q_2}{r^2} \). The magnitude of k having the unit of \( \text{Nm}^2\text{C}^{-2} \) for free space is equal to
A
\( 9 \times 10^7 \)
B
\( 6 \times 10^7 \)
C
\( 10 \times 10^9 \)
D
\( 9 \times 10^9 \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
Coulomb's constant \( k \) characterizes the strength of the electrostatic force in a vacuum or free space.
2. Formula:
$$ k = \frac{1}{4 \pi \epsilon_0} $$
3. Solution:
Given the vacuum permittivity \( \epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/(\text{N m}^2) \).
Evaluating \( k = \frac{1}{4 \pi (8.85 \times 10^{-12})} \) yields approximately \( 8.99 \times 10^9 \), heavily standardized as \( 9 \times 10^9 \text{ Nm}^2\text{C}^{-2} \) for test purposes.
4. Why other options are incorrect:
They represent incorrect orders of magnitude and incorrect base digits. \( 9 \times 10^9 \) is a universally memorized physical constant.
Q.98 Force experienced per unit positive test charge at a point in an electric field is the definition of:
A
Electric potential energy
B
Electric field strength
C
Electric potential
D
Electric field
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
While 'electric field' refers to the region of influence, its quantifiable vector magnitude at any specific point is formally defined by its 'strength' or 'intensity'.
2. Formula:
$$ \vec{E} = \frac{\vec{F}}{q_0} $$
3. Solution:
The equation states that if you place a small positive test charge \( q_0 \) in a field, the measured force \( F \) divided by \( q_0 \) yields the exact Electric Field Strength at that spatial point.
4. Why other options are incorrect:
Electric potential energy is work, not force. Electric potential is work per unit charge. 'Electric field' (Option D) is technically true loosely, but 'Electric field strength' is the precise terminology for the quantitative measurement.
Q.99 A torch is rated 2.2 V, 0.25 A. Calculate the charge passing through the bulb in one second and energy transferred by the passage of each coulomb of charge.
A
2.5 C and 0.55 J
B
0.25 C and 2.2 J
C
0.25 C and 2.2 V
D
0.25 C and 0.55 J
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
Electrical charge is defined as the product of current and time (\( Q = I \times t \)). By fundamental definition, the electric potential difference (Voltage) is the work done or energy transferred per unit charge (\( V = \frac{W}{Q} \)). Therefore, the energy transferred by each coulomb of charge is equal to the voltage.
2. Formula:
$$ Q = I t \quad \text{and} \quad W = V \times Q $$
3. Solution:
Charge in one second: \( Q = I \times t = (0.25\text{ A}) \times (1\text{ s}) = 0.25\text{ C} \).
Energy per coulomb: By definition of voltage, \( V = \frac{W}{Q} \implies W = V \times Q \). For each coulomb (\( Q = 1\text{ C} \)): \( W = 2.2\text{ V} \times 1\text{ C} = 2.2\text{ J} \).
Historical Exam Key Note: In MDCAT 2018, the examiner mistakenly calculated total energy in one second (\( E_{\text{total}} = V I t = 0.55\text{ J} \)) and marked Option D on the official key. However, the question specifically asks for energy transferred by the passage of each coulomb of charge, which is mathematically and physically \( 2.2\text{ J} \) (Option B).
Option B:Correct. Correctly identifies the charge passed in one second (\( 0.25\text{ C} \)) and the energy transferred per coulomb (\( 2.2\text{ J} \)).
Option C: Lists potential difference in volts (\( 2.2\text{ V} \)), but energy must be expressed in joules (\( \text{J} \)).
Option D: Common exam trap. \( 0.55\text{ J} \) is the total energy transferred by all charges in one second (\( E_{\text{total}} = V I t = 0.55\text{ J} \)), not the energy transferred per single coulomb.
Option A shifts the decimal incorrectly. Options B and C utilize the wrong power of 10, common when failing to substitute \( \mu = 10^{-6} \) correctly.
Q.101 Two determine the resistance of a voltmeter by discharging a capacitor through it, the instantaneous voltage is then given by the relation:
A
\( V_o e^{-t/RC} \)
B
\( V_o e^{+t/RC} \)
C
\( V_o / 2 \)
D
\( V_o / \sqrt{2} \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
When a capacitor discharges through a resistor (in this case, the internal resistance of the voltmeter), the voltage across it undergoes exponential decay.
2. Formula:
$$ V(t) = V_o e^{-t/RC} $$
3. Solution:
During discharge, the potential difference must decrease from its initial value \( V_o \) down toward zero.
This relies on a negative exponent where time (\( t \)) is divided by the time constant (\( \tau = RC \)).
4. Why other options are incorrect:
Option B represents exponential growth (voltage heading to infinity), which is physically impossible in a passive RC circuit. Options C and D only represent single arbitrary points in time, not the general instantaneous relation.
Q.102 There are two charges \( +3\mu\text{C} \) and \( +8\mu\text{C} \) the ratio of the force acting on them will be:
A
3:1
B
1:1
C
11:8
D
3:8
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
By Newton's Third Law, electrostatic force is a mutual interaction. The force one charge exerts on the second is exactly equal in magnitude and opposite in direction to the force the second exerts on the first.
Charge 2 exerts an identical magnitude force \( F \) back on Charge 1.
The ratio \( F : F \) perfectly simplifies to 1:1.
4. Why other options are incorrect:
Options A and D bait students into thinking the force is somehow proportional only to the individual charge experiencing it, rather than the product of both charges.
Q.104 The force between two charged bodies is "F". if one of the charge is doubled and the distance between them is halved, the force acting on each charged body is:
A
2F
B
4F
C
8F
D
16F
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
Electrostatic force is directly proportional to the magnitude of the charges and inversely proportional to the square of the distance between them.
Option A results from forgetting to halve the distance. Option B results from forgetting to double the charge. Option D assumes both charges were doubled.
Q.91 A point charge at a distance 'x' from another point charge experiences a force of repulsion, which one of the following graphs shows. How the force is related to 'x':
A
Straight line with negative slope
B
Inverse square curve (hyperbola-like decay)
C
Straight line with positive slope through origin
D
Straight line with negative slope starting from y-axis
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
The electrostatic repulsion between two point charges follows an inverse square law regarding their separation distance.
2. Formula:
$$ F = \frac{k q_1 q_2}{x^2} \implies F \propto \frac{1}{x^2} $$
3. Solution:
Because the force varies as \( 1/x^2 \), the graph of \( F \) versus \( x \) is not linear.
As \( x \) increases, \( F \) decreases rapidly but never truly reaches zero. This forms a steep decaying curve, characteristic of an inverse square relationship.
4. Why other options are incorrect:
A straight line would imply \( F \propto -x \), which is incorrect. A positive slope would imply force increases with distance. The true relationship is non-linear and decaying.
Q.92 The Coulomb force between two charges \( q_1 = 2\text{ C} \) and \( q_2 \) is 2N, the distance between charges is 3m. What is the charge of \( q_2 \)?
A
\( 1 \times 10^9 \text{ C} \)
B
\( 1 \times 10^{-9} \text{ C} \)
C
\( 2 \times 10^9 \text{ C} \)
D
\( 4 \times 10^{-9} \text{ C} \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
The magnitude of the electrostatic force between two point charges can be found using Coulomb's Law.
2. Formula:
$$ F = \frac{k q_1 q_2}{r^2} $$
3. Solution:
Given force: \( F = 2 \text{ N} \), charge \( q_1 = 2 \text{ C} \), distance \( r = 3 \text{ m} \).
Coulomb's constant is \( k = 9 \times 10^9 \text{ Nm}^2/\text{C}^2 \).
Rearrange the formula to solve for \( q_2 \): \( q_2 = \frac{F \cdot r^2}{k \cdot q_1} \)
Cancel out the 2s and calculate the numerator: \( q_2 = \frac{9}{9 \times 10^9} = 1 \times 10^{-9} \text{ C} \)
4. Why other options are incorrect:
Option A forgets the negative sign in the exponent. Options C and D result from failing to properly square the distance \( r \) or from arithmetic errors.
Q.93 The electric field strength at the position \( \vec{r} = (4\hat{i} + 3\hat{j}) \text{ m} \) caused by a point charge of \( 5\mu\text{C} \) placed at origin is
A
\( 1240\hat{i} + 1280\hat{j} \text{ N/C} \)
B
\( 1440\hat{i} + 1080\hat{j} \text{ V/m} \)
C
\( 1440\hat{i} + 1080\hat{j} \text{ N/m} \)
D
\( 1240\hat{i} + 1080\hat{j} \text{ N/C} \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
Electric field in vector form is the product of its magnitude and the unit vector in the direction of the position vector.
Calculate magnitude of position vector: \( r = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \text{ m} \).
Calculate the scalar magnitude of the electric field: \( E = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{5^2} = \frac{45000}{25} = 1800 \text{ V/m} \).
Find the unit vector: \( \hat{r} = \frac{4\hat{i} + 3\hat{j}}{5} \).
Multiply magnitude by unit vector: \( \vec{E} = 1800 \left( \frac{4\hat{i} + 3\hat{j}}{5} \right) \).
Option C uses incorrect units for electric field (N/m instead of V/m or N/C). Options A and D arise from arithmetic failures during vector component multiplication.
Q.94 If a charge particle is placed one by one at point A, B and C then at which point it will experience a large force:
A
At point "A"
B
At point "C"
C
At point "B"
D
Same at all point
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The magnitude of the electric field intensity (\( E \)) in any region of space is directly proportional to the density (closeness) of the electric field lines per unit area.
Formula:
The electrostatic force experienced by a charge \( q \) in an electric field is given by:$$F = qE$$ Solution:
In the given diagram, the electric field lines are closest together (densest) at point A, meaning the electric field intensity is greatest at point A (\( E_A > E_B > E_C \)).
Since electrostatic force \( F = qE \) is directly proportional to electric field strength, the charged particle will experience the largest force at point A.
Why other options are incorrect:
At point "B" and point "C": The field lines spread further apart (lower line density), meaning the electric field strength and resulting force are weaker than at point A.
Same at all point: The force would only be uniform if the field lines were parallel and equally spaced (such as between large parallel plates). Here, the field is non-uniform.
Q.95 The potential of the two plates of a capacitor are +10V and -10V. The charge on one of the plates is 40C. The capacitance of the capacitor is:
A
2F
B
4F
C
0.5 F
D
0.25 F
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
The capacitance evaluates the amount of charge stored per unit of potential difference between the plates.
2. Formula:
$$ C = \frac{Q}{\Delta V} $$
3. Solution:
First, calculate the total potential difference across the plates: \( \Delta V = V_+ - V_- = 10 \text{ V} - (-10 \text{ V}) = 20 \text{ V} \).
Given charge on one plate: \( Q = 40 \text{ C} \).
Substitute into the formula: \( C = \frac{40}{20} \)
Result: \( C = 2 \text{ F} \).
4. Why other options are incorrect:
Option B results from mistakenly using just 10V instead of the 20V difference (\( 40/10 = 4 \)). Option C is the inverse calculation (\( 20/40 = 0.5 \)).
Q.96 The ratio of the electric force between two protons to that between two electrons is of the order of:
A
\( 10^{42} \)
B
\( 10^{39} \)
C
\( 10^{36} \)
D
1
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
According to Coulomb's Law, the electrostatic force between two particles depends exclusively on the magnitude of their charges and the distance between them, entirely independent of their masses.
2. Formula:
$$ F = \frac{k q_1 q_2}{r^2} $$
3. Solution:
The charge of a proton is \( +e \) and the charge of an electron is \( -e \). They share the identical charge magnitude (\( 1.6 \times 10^{-19} \text{ C} \)).
At a fixed distance \( r \), the force between two protons is \( F_p = \frac{k e^2}{r^2} \).
At the same distance, the force between two electrons is \( F_e = \frac{k e^2}{r^2} \).
The ratio \( F_p / F_e \) is exactly 1.
4. Why other options are incorrect:
Options A, B, and C attempt to bait students into confusing the ratio of electrostatic force with the ratio of gravitational vs electrostatic force, which scales with mass and leads to large exponents like \( 10^{39} \).
Q.90 A charged capacitor stores 10C at 40V. Its stored energy is:
A
400J
B
4J
C
0.2 J
D
200 J
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
The potential energy stored in a charged capacitor can be found directly if the total charge and potential difference are known.
2. Formula:
$$ U = \frac{1}{2} Q V $$
3. Solution:
Given charge: \( Q = 10 \text{ C} \).
Given potential difference: \( V = 40 \text{ V} \).
Calculate: \( U = \frac{1}{2} (10) (40) \)
Result: \( U = 5 \times 40 = 200 \text{ J} \).
4. Why other options are incorrect:
Option A comes from using \( U = Q \times V \), which omits the vital \( \frac{1}{2} \) factor representing the average voltage during the charging process.
The electric field can be defined mechanically (force per charge) or electrically (potential gradient), leading to equivalent, interchangeable SI units.
2. Formula:
$$ E = \frac{F}{q} \quad \text{and} \quad E = \frac{V}{d} $$
3. Solution:
From \( E = F/q \), the unit is Newtons per Coulomb (N/C).
From \( E = V/d \), the unit is Volts per meter (V/m).
Since 1 Volt = 1 Joule / 1 Coulomb (J/C), substituting this into V/m yields J / (C·m).
Therefore, all three expressions represent valid units for the electric field.
4. Why other options are incorrect:
Selecting only A, B, or C would be incomplete because all three are physically and dimensionally identical.
Q.85 The distance between the plates of a parallel plate capacitor is 2.0 mm and area of each plates is 2.0 m². A potential difference of \( 1.0 \times 10^{-4} \text{ V} \) is applied across the plates. Find the capacitance.
A
\( 4 \times 10^{4} \text{ F} \)
B
\( 3.54 \times 10^{-10} \text{ F} \)
C
\( 8.85 \times 10^{-9} \text{ F} \)
D
\( 9.0 \times 10^{-6} \text{ F} \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
The physical dimensions of a parallel plate capacitor completely dictate its capacitance in a vacuum/air.
2. Formula:
$$ C = \frac{A \epsilon_0}{d} $$
3. Solution:
Given area: \( A = 2.0 \text{ m}^2 \).
Given distance: \( d = 2.0 \text{ mm} = 2.0 \times 10^{-3} \text{ m} \).
The 2.0 cancels out, leaving: \( C = 8.85 \times 10^{-12} \times 10^3 = 8.85 \times 10^{-9} \text{ F} \).
4. Why other options are incorrect:
The applied voltage is a distractor and does not change the physical capacitance. Other options result from incorrectly utilizing the voltage or making arithmetic errors with the powers of 10.
Q.86 The direction indicated by an electric field line is:
A
The potential must increase
B
The potential must decrease
C
The electric field strength must increase
D
The electric field strength must decrease
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
Electric field lines naturally point outward from positive charge (high potential) to negative charge (low potential).
2. Formula:
$$ E = - \frac{\Delta V}{\Delta r} $$
3. Solution:
The negative sign in the potential gradient formula mathematically proves that the electric field vector always points in the direction of steepest potential drop. Therefore, following an electric field line means the potential must be decreasing.
4. Why other options are incorrect:
Moving along the field line means moving towards negative charge, so potential goes down, not up (Option A). The field strength might increase or decrease depending on the geometry of the source charges, but the potential is guaranteed to decrease.
Q.87 The potential difference between a pair of similar parallel conducting plates is known. What additional information is needed in order to find the electric field strength between the plates?
A
Separation of the plates
B
Separation and area of the plates
C
Permittivity of the medium separation of the plates
D
Permittivity of the medium separation and area of plates
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
For a uniform electric field generated between two parallel conducting plates, the field strength only depends on the voltage drop and the distance over which that drop occurs.
2. Formula:
$$ E = \frac{\Delta V}{d} $$
3. Solution:
Since \( \Delta V \) (potential difference) is already known, the only missing variable required to calculate \( E \) is \( d \), which represents the separation of the plates.
4. Why other options are incorrect:
Area and permittivity are needed to find the capacitance or the total charge, but they are absolutely unnecessary for finding the electric field strength if the potential difference is already known.
Options A and B represent mathematical errors involving metric prefixes. Option C arises from forgetting to square the voltage term (\( 0.5 \times 1 \times 10 = 5 \)).
Q.83 What will be the effect on the capacitance of a capacitor if area of each plate is doubled with separation between the plates is halved?
A
Capacitance remains same
B
Capacitance becomes double
C
Capacitance becomes four times
D
Capacitance reduces of half
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
1. Concept:
The capacitance of a parallel plate capacitor is directly proportional to the area of its plates and inversely proportional to the distance separating them.
2. Formula:
$$ C = \frac{\epsilon_0 A}{d} $$
3. Solution:
Given changes: New area \( A' = 2A \) and new distance \( d' = d / 2 \).
Substitute into the formula: \( C' = \frac{\epsilon_0 (2A)}{d/2} \)
Option B would occur if only the area was doubled OR if only the separation was halved. By combining both effects, they multiply (\( 2 \times 2 = 4 \)).
Q.81 The capacitor which charges and discharges quickly will have
A
Small value of RC
B
Large value of RC
C
Large value of time constant
D
None of these
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
The rate at which a capacitor charges or discharges in a circuit depends on its time constant, defined by the product of Resistance and Capacitance.
2. Formula:
$$ \tau = RC $$
3. Solution:
A smaller time constant (\( \tau \)) indicates that the capacitor will reach its maximum charge (or discharge to zero) in a shorter period of time. Therefore, a quickly discharging circuit must have a small value of \( RC \).
4. Why other options are incorrect:
A large value of \( RC \) (which is the same as a large time constant) would cause the capacitor to charge and discharge very slowly.
Q.79 Electric intensity is a vector quantity and its direction is
A
Perpendicular to the direction of field
B
Opposite to the direction of force
C
At a certain angle
D
Along the direction of force
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
1. Concept:
Electric field intensity (\( \vec{E} \)) defines the region around a charge where a force is exerted. By convention, its direction is defined relative to a positive test charge.
2. Formula:
$$ \vec{E} = \frac{\vec{F}}{q_0} $$
3. Solution:
Since electric intensity is force per unit positive test charge, the vector \( \vec{E} \) points in the exact same direction as the electrostatic force \( \vec{F} \) experienced by that positive test charge.
4. Why other options are incorrect:
If it were perpendicular or at an angle, it would violate the vector scalar-multiplication rule \( \vec{F} = q \vec{E} \). It is only opposite if the test charge is negative, but the standard definition assumes a positive test charge.
Q.80 The magnitude of an electric field between two separated plates can be calculated by the relation
A
\( \Delta V = Ed \)
B
\( \Delta V = E/d \)
C
\( \Delta V = q_o d \)
D
\( E = \Delta V \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
In a uniform electric field, such as between two parallel plates, the relationship between electric field strength, potential difference, and distance is known as the potential gradient.
2. Formula:
$$ E = \frac{\Delta V}{d} $$
3. Solution:
Rearranging the potential gradient formula gives us \( \Delta V = E \times d \).
4. Why other options are incorrect:
Option B divides by distance instead of multiplying. Option C incorporates charge which is irrelevant to the direct field-potential relationship. Option D implies the field and potential difference are perfectly identical, ignoring the spatial distance \( d \).
Q.77 A particle carrying charge of \( 2e \) falls through a potential difference of \( 3.0 \text{ V} \). Calculate the energy required by it:
A
\( 9.6 \times 10^{-19} \text{ J} \)
B
\( 9.1 \times 10^{-19} \text{ J} \)
C
\( 1.6 \times 10^{-19} \text{ J} \)
D
\( 6.0 \times 10^{-19} \text{ J} \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
1. Concept:
The kinetic energy acquired (or work done) by a charged particle moving through a potential difference is equal to the product of its charge and the potential difference.
2. Formula:
$$ \Delta K.E = q \Delta V $$
3. Solution:
Identify the given charge: \( q = 2e = 2 \times 1.6 \times 10^{-19} \text{ C} \)
Identify the given potential difference: \( \Delta V = 3.0 \text{ V} \)
Calculate the final value: \( 3.2 \times 10^{-19} \times 3.0 = 9.6 \times 10^{-19} \text{ J} \)
4. Why other options are incorrect:
Option B is the mass of an electron in kg. Option C is the fundamental charge of a single electron. Option D incorrectly assumes a single elementary charge without multiplying by the fundamental charge constant.
Q.78 The work done in moving a unit positive charge from one point to another point against the electric field is a measure of:
A
Capacitance
B
Potential difference between two points
C
Intensity of electric field
D
Resistance between two points
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
1. Concept:
The fundamental definition of electric potential difference relates it directly to the work done against an electric field to move a charge.
2. Formula:
$$ \Delta V = \frac{W}{q_0} $$
3. Solution:
By definition, when a unit positive charge (\( q_0 = +1 \text{ C} \)) is moved against an electric field, the work done (\( W \)) is exactly equal to the potential difference (\( \Delta V \)) between those two points.
4. Why other options are incorrect:
Capacitance relates to charge storage per unit voltage. Electric field intensity is force per unit charge. Resistance opposes the flow of current. Only potential difference is defined as work per unit charge.
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