Physics Electrostatics MDCAT 2019
PMDC Verified Question 102 of 131
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
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: \( 125 \times 1.6 \times 10^{-19} \text{ J} \)
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.

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