Physics Electrostatics ETEA 2022
PMDC Verified Question 73 of 131
Q.133 The energy stored in a capacitor is given by:
A
\( U = \frac{1}{2} QV^2 \)
B
\( U = \frac{1}{2} CV^2 \)
C
\( U = \frac{1}{2} QC \)
D
\( U = \frac{1}{2V} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: \( U = \frac{1}{2} CV^2 \)
1. Concept:

The work done to build up charge on a capacitor is stored as electric potential energy. This can be expressed in terms of Capacitance and Voltage.

2. Formula:

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

3. Solution:

  • Because voltage increases linearly as charge is added (\( V = Q/C \)), the average voltage during charging is \( \frac{1}{2} V \).


  • The total energy is Work = Charge \( \times \) Average Voltage = \( Q \times (\frac{1}{2} V) = \frac{1}{2} Q V \).


  • Substituting \( Q = CV \) yields \( U = \frac{1}{2} (CV) V = \frac{1}{2} C V^2 \).


4. Why other options are incorrect:

Option A incorrectly squares the voltage while keeping charge. Option C multiplies charge and capacitance, which is dimensionally invalid for energy.

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