Physics Electrostatics DUHS 2022
PMDC Verified Question 81 of 131
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
\( 32 \times 10^{19} \text{ J} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: \( 32 \times 10^{-19} \text{ J} \)
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.

2. Formula:

$$ E = q \Delta V $$

3. Solution:

  • Given charge: \( q = 1.6 \times 10^{-19} \text{ C} \).


  • Given potential difference: \( V = 20 \text{ V} \).


  • Calculate energy: \( E = (1.6 \times 10^{-19}) \times 20 \).


  • Result: \( E = 32 \times 10^{-19} \text{ J} \).


  • 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.

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