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

d' = d /2

Step-by-step explanation:

Given that

Distance = d

Voltage =V

We know that energy in capacitor given as

`U=(1)/(2)CV^2`

`C=(\varepsilon _oA)/(d)`

`U=(1)/(2)* (\varepsilon _oA)/(d)* V^2`

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

`U'=(1)/(2)* (\varepsilon _oA)/(d')* V^2`

`2U=(1)/(2)* (\varepsilon _oA)/(d')* V^2`

`2* (1)/(2)* (\varepsilon _oA)/(d)* V^2=(1)/(2)* (\varepsilon _oA)/(d')* V^2`

2 d ' = d

d' = d /2

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