The left plate of the capacitor shown in the figure above carries a charge +Q while the right plate is uncharged at t=0 . The total charge on the right plate after closing the switch will be

Step-by-step solution

Electric field between the capacitor plates = ₁ 2 ₀ + (- ₂ ) 2 ₀ E = Q + X 2 ~A ₀ + X 2 ~A ₀ = 1 2 ~A ₀ [ Q +2 x ] Potential different E _ d = d 2 A ₀ [Q+2 x]= = Q +2 x 2 C - x = Q 2 - C