AP EAMCET202316 May 2023Morning ShiftPhysicsCapacitanceActual
Between the plates of a parallel plate capacitor of plate area A and capacity 0.025 F , a metal plate of area, A and thickness equal to 1 3 of the separation between the plates of the capacitor is introduced. If the capacitor is charged to 100 ~V , then the amount of work done to remove the metal plate from the capacitor is
Options
- A62.5 J
- B30.2 J
- C52.6 J
- D35.4 J
Correct answer
A. 62.5 J
Step-by-step solution
Initially c^ = A d- d 3 = 3 2 ₀ A d = 3 2 c [C=capacitor of plate without conduction] aligned & v_i=v_f=100 v & u_i= 1 2 c v^2 & u_f= 1 2 c v^2 aligned v_i-v_f= 1 2 ( 3 2 c-c ) v^2= 1 2 c 2 v^2 aligned & = 0.025 10^4 4 J & = 250 4 =62.5 J aligned