Question

A parallel-plate capacitor has plates of area A. The plates are initially separated by a distance d, but this distance can be varied. The capacitor is connected to a battery.What should the plate separation be if you want to halve the energy density?

Answer 1

Answer:

d' = d /2

Explanation:

Given that

Distance = d

Voltage =V

We know that energy in capacitor given as

$U=\dfrac{1}{2}CV^2$

$C=\dfrac{\varepsilon _oA}{d}$

$U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2$

If energy become double U' = 2 U then d'

$U'=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2$

$2U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2$

$2\times \dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2$

2 d ' = d

d' = d /2

So the distance between plates will be half on initial distance.