# Parallel plate capacitor?

• Jun 29th 2009, 01:40 PM
fardeen_gen
Parallel plate capacitor?
A parallel-plate capacitor has plates of area A and separation d and is charged to a potential difference V. The charging battery is then disconnected, and the plates pulled apart until their separation is 2d. Derive expression in terms of A, d, and V for:

(a) the new potential difference
(b) the initial and final stored energies, $\displaystyle U_i$ and $\displaystyle U_f$
(c) the work required to separate the plates

Spoiler:

$\displaystyle \mbox{(a)}2V$
$\displaystyle \mbox{(b)}\frac{\epsilon_0 AV^2}{2d}$
$\displaystyle \mbox{(c)}\frac{\epsilon_0 AV^2}{2d}\ \frac{\epsilon_0 AV^2}{2d}\ V = L[\sqrt{y^2 + L^2 - y}]\ E_g = \frac{1}{4\pi \epsilon_0}\left[1 - \frac{y}{\sqrt{y^2 + L^2}}\right]$

How to do it?
• Jun 30th 2009, 04:28 AM
malaygoel
The charge on the capacitor will remain constant.

and Capacitance will vary as $\displaystyle \frac{\epsilon_0 A}{d}$

hence capacitance will decrease by a factoor of 2.

$\displaystyle Q=CV$

hence, Potential between the plates will increase by a factor of 2.