A parallel plate capacitor with area 200 ~cm ^2 and separation between the plates 1.5 ~cm , is connected across a battery of emf V. If the force of attraction between the plates is 25 10⁻⁶ ~N , the value of V is approximately: . ( ₀=8.85 10⁻¹² C ^2 N.m )^2 )

Step-by-step solution

Given area of Parallel plate capacitor, A=200 ~cm ^2 Separation between the plates, d=1.5 ~cm Force of attraction between the plates, F=25 10⁻⁶ ~N F=Q EF= Q^2 2 A ₀ (E due to parallel plate .= 2 ₀ = Q A 2 ₀ ) But Q=C V= ₀ A(V) d gathered . F= ( ₀ A V^2 ) d ^2 2 ~A ₀ ) = ( ₀ A )^2 V^2 d^2 2 (A ₀ ) = ( ₀ A ) V^2 d^2 2 or, 25 10⁻⁶= (8.85 10⁻¹² ) (200 10⁻⁴ ) V^2 2.25 10⁻⁴ 2 V= 25 10⁻⁶ 2.25 10⁻⁴ 2 8.85 10⁻¹² 200 10⁻⁴ 250 ~V gathered