Physics Electrostatics MDCAT 2017
PMDC Verified Question 113 of 131
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
\( 4 \times 10^{-9} \text{ C} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: \( 1 \times 10^{-9} \text{ C} \)
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 \( q_2 \): \( q_2 = \frac{F \cdot r^2}{k \cdot q_1} \)


  • Substitute values: \( q_2 = \frac{2 \times 3^2}{(9 \times 10^9) \times 2} \)


  • 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.

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