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ipn [44]
3 years ago
11

A parallel-plate capacitor has plates of area A. The plates are initially separated by a distance d, but this distance can be va

ried. The capacitor is connected to a battery.What should the plate separation be if you want to halve the energy density?
Physics
1 answer:
ohaa [14]3 years ago
7 0

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.

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