Answer
The potential difference between the plates is $~~16.0~V$
Work Step by Step
We can find the charge on the capacitor:
$q = C~V$
$q = (\frac{\epsilon_0~A}{d})(V)$
$q = \frac{(8.854\times 10^{-12}~C/V~m)(8.50\times 10^{-4}~m^2)(6.00~V)}{3.00\times 10^{-3}~m}$
$q = 1.505\times 10^{-11}~C$
We can find the potential difference between the plates after they are pulled apart:
$V = \frac{q}{C}$
$V = \frac{q}{\epsilon_0~A/d}$
$V = \frac{q~d}{\epsilon_0~A}$
$V = \frac{(1.505\times 10^{-11}~C)(8.00\times 10^{-3}~m)}{(8.854\times 10^{-12}~C/V~m)(8.50\times 10^{-4}~m^2)}$
$V = 16.0~V$
The potential difference between the plates is $~~16.0~V$.
