Physics (10th Edition)

$1.1\times10^{3}$ V
Recall: U= $\frac{1}{2}CV^{2}$. Rearranging the equation, we get: V= $\sqrt {\frac{2U}{C}}$ Plugging in the values, we have: Potential difference: V= $\sqrt {\frac{2\times73J}{120\times10^{-6}F}}= 1.1\times10^{3}V$