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
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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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