## College Physics (7th Edition)

(a) $q_{1}=CV_{1}$ $q_{2}=CV_{2}$ Given that $V_{2}=2V_{1}$. Therefore, $q_{2}=C(2V_{1})=2CV_{1}=2q_{1}$ If the potential difference is doubled, charge also gets doubled. (b) $E_{1}=\frac{1}{2}CV_{1}^{2}$ $E_{2}=\frac{1}{2}CV_{2}^{2}$ Given that $V_{2}=2V_{1}$. Therefore, $E_{2}=\frac{1}{2}C(2V_{1})^{2}=4(\frac{1}{2}CV_{1}^{2})=4E_{1}$. Energy increases by a factor of 4.