## College Physics (4th Edition)

$\Delta V = 8200~V$
We can find the potential difference: $U = \frac{1}{2}C~(\Delta V)^2$ $(\Delta V)^2 = \frac{2~U}{C}$ $\Delta V = \sqrt{\frac{2~U}{C}}$ $\Delta V = \sqrt{\frac{(2)(300~J)}{9\times 10^{-6}~F}}$ $\Delta V = 8200~V$