Answer
(a) $\Delta V = 3000~V$
(b) $Q = 6.0\times 10^{-3}~C$
Work Step by Step
(a) We can find the maximum potential difference before breakdown:
$\Delta V = E~d$
$\Delta V = (3\times 10^6~V/m)(0.0010~m)$
$\Delta V = 3000~V$
(b) We can find the magnitude of the greatest charge the capacitor can store:
$Q = C~\Delta V$
$Q = (2.0\times 10^{-6}~F)(3000~V)$
$Q = 6.0\times 10^{-3}~C$
