First slide

Potential and charge distribution for

Question

An alternating voltage $v = 100 \cos 100 π t$ volt is applied across a $100 μ F$ capacitor. Then average electrostatic energy stored in it is

Moderate

A

zero
B

1.5 J
C

0.5 J
D

0.25 J

Solution

$\begin{array}{l} \frac{Q}{C} = V \Rightarrow \frac{Q}{100 \times 10^{- 6}} = 100 \cos 100 π t \\ \Rightarrow Q = 10^{- 2} \cos 100 π t coulomb \end{array}$
$\begin{array}{l} ∴ Q_{r m s} = \frac{10^{- 2}}{\sqrt{2}} coulomb \\ ∴ U = Q_{r m s}^{2} / 2 C = {(\frac{10^{- 2}}{\sqrt{2}})}^{2} \times \frac{1}{2 \times 10^{- 4}} J = 0.25 J \end{array}$

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