Physics Electrostatics PMC 2021
PMDC Verified Question 94 of 131

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
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: 4 kV
Concept:

Calculate potential difference from electrostatic potential energy stored in a capacitor's electric field.

Formula:

$$U = \frac{1}{2} C V^2$$

Solution:

Given values:
  • Energy stored \( U = 24\text{ J} \)
  • Capacitance \( C = 3\mu\text{F} = 3 \times 10^{-6}\text{ F} \)


Rearrange to solve for voltage \( V \):

$$24 = \frac{1}{2} \times (3 \times 10^{-6}) \times V^2$$

$$48 = (3 \times 10^{-6}) \times V^2$$

$$V^2 = \frac{48}{3 \times 10^{-6}} = 16 \times 10^6$$

$$V = \sqrt{16 \times 10^6} = 4 \times 10^3\text{ V} = 4\text{ kV}$$

Why other options are incorrect:

  • 16 kV is the value of \( V^2 \) without taking the square root.


  • Other values arise from simple algebraic mistakes during division or multiplication.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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