Physics Electrostatics MDCAT 2009
PMDC Verified Question 130 of 131
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
\( 6.0 \times 10^{-19} \text{ J} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: \( 9.6 \times 10^{-19} \text{ J} \)
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: \( \Delta V = 3.0 \text{ V} \)


  • Substitute the values: \( \Delta K.E = (2 \times 1.6 \times 10^{-19} \text{ C}) \times (3.0 \text{ V}) \)


  • 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.

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