#### Answer

a) charge gets doubled.
b) increases by a factor of 4.

#### Work Step by Step

(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.