Two parallel plate capacitors 8 F each are connected in parallel to a 10 ~V battery. The plate separation in one of the capacitor is reduced to 40 % of its initial value. The increase in the total charge stored on the capacitors is

Step-by-step solution

Given, aligned C₁ & =C₂=8 F =8 10⁻⁶ ~F V & =10 ~V aligned Since, C₁ and C₂ are connected in parallel. Equivalent capacitance, aligned C_ eq =C₁+C₂= & 8 10⁻⁶+8 10⁻⁶=1.6 10⁻⁵ ~F Total charge, q_i & =C_ eq V & =1.6 10⁻⁵ 10 & =1.6 10⁻⁴ C aligned We know that, capacitance C= ₀ A d array ll & C 1 d & C^ C^ = d^ d^ = (d^ -40 % of d^ ) d^ = 3 / 5 d^ d^ array aligned & C^ C^ = 3 5 & C^ = 5 3 C^ & = 5 3 8 10⁻⁶= 40 3 10⁻⁶ C & aligned New capacitance, C^ =C^ +C₂ aligned & = 40 3 10⁻⁶+8 10⁻⁶ & = 64 3 10⁻⁶ ~F aligned Now value o