Question

A capacitor stores 7.77 x 10-7 J of

Answer 1

Answer: $36.22\ V$

Explanation:

Given

Energy in a capacitor $E=7.77\times 10^{-7}\ J$

Charge $Q=4.29\times 10^{-8}\ C$

Energy of a capacitor is given by

$E=\dfrac{1}{2}CV^2=\dfrac{1}{2}QV\quad [\text{Q=CV}]$

Insert the values

$E=\dfrac{1}{2}QV\\\\\Rightarrow 7.77\times 10^{-7}=\dfrac{1}{2}\times 4.29\times 10^{-8}\times V\\\\\Rightarrow V=3.622\times 10\\\\\Rightarrow V=36.22\ V$

Thus, the voltage around the capacitor is $36.22\ V$