Physics, asked by Philipshooter03, 9 months ago

A parallel plate capacitor having plate area 'A' , and separation between the plates 'd' is charged to potential V . The energy stored in the capacitor​

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Answers

Answered by amitnrw
6

E = (1/2)ε₀(V²/d) A

Explanation:

E = The energy stored in the capacitor​

E = (1/2)CV²

Where C = Aε₀/d

Substituting C = Aε₀/d

in E = (1/2)CV²

=> E = (1/2)(Aε₀/d)V²

=> E = (1/2)ε₀(V²/d) A

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Answered by Anonymous
3

Given :

Area of plate = A

Separation between the plates = d

Potential = V

To find :

The energy stored in the capacitor .

Solution :

The energy stored in the capacitor​  , E = ( 1 / 2 ) C V²     ....(i)

Capacitance C of a capacitor having plate area 'A' , and separation between the plates 'd' is given by ,

=> C = Aε₀/d

Substituting C = Aε₀/d  in (i)

    E = ( 1 / 2 ) * ( A * ε₀ / d ) * V²

=> E = ( 1 / 2 ) * ε₀ * ( V² / d ) * A

The energy stored in the capacitor is ( 1 / 2 ) * ε₀ * ( V² / d ) * A

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