Four uncharged metal plates of the same area are placed in vacuum and separated by equal distances. They are connected by conducting wires to two resistors, two open switches and a battery. The thickness of the plates is negligible, and their area is sufficiently large. S1 is now closed, and after sufficient time elapses, a charge with quantity Q is stored in A. Next, as S1 is kept closed, S2 is also closed. After sufficient time elapses, a charge with quantity of electricity Q' is stored in A. What is the ratio of Q'/Q?
Q = C.V
C = ε0 . A / d
The Attempt at a Solution
I think the circuit acts as capacitor.
When S1 is closed and S2 is open, the circuit is like capacitor between plate A and D because potential difference only applied to the two plates by the battery. Assume the distance between two adjacent plates is d, then the distance of AD = 3d. So the charge stored on A = ε0 . A / 3d . V
When both switches closed, I don't have idea how the circuit works....