Answer: [tex]36.22\ V[/tex]
Explanation:
Given
Energy in a capacitor [tex]E=7.77\times 10^{-7}\ J[/tex]
Charge [tex]Q=4.29\times 10^{-8}\ C[/tex]
Energy of a capacitor is given by
[tex]E=\dfrac{1}{2}CV^2=\dfrac{1}{2}QV\quad [\text{Q=CV}][/tex]
Insert the values
[tex]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[/tex]
Thus, the voltage around the capacitor is [tex]36.22\ V[/tex]
