The plates of a parallel-plate capacitor have surface area A and separation dist

Question

The plates of a parallel-plate capacitor have surface area A and separation distance d. The capacitor (Co=EoA/d) is connected to a battery and is charged completely to charge Q. After the capacitor is disconnected from the battery, the plates are separated a bit so that a neutral conducting slab can be inserted between them to create a three-plate capacitor, as shown in the lower figure. The surface area of the slab is also A, and the thickness of the slab is h.

(a) Once the slab is inserted, electrons in the slab quickly arrange themselves so that E=0 inside the slab. On the lower figure, sketch and label the equilibrium charge distribution for the slab.

(b) Calculate C, the capacitance of the three-plate capacitor. Do this by first integrating to find V, the potential difference between the uppermost and lowermost plates on the capacitor.

(c) Calculate C(eq) for two parallel plate capacitors (each with area A and separation d) that are connected in series with one another. Compare C(eq) to your answer from part (b).

Explanation / Answer

You need to add volatages on the resistor and on the capacitor, taking into account their relative phase, which is 90 degrees. Current in resistor and capasitor connceted in series is the same, lets denote this current I. Impedance of the resistor Z(R) is R. Impedance of the capacitor Z(C) is 1/(i?C), where i = v-1. Combined votage is V = I x Z(R) + I Z(C) = I (R + 1/(i?C)) V/I = R + 1/(i?C) (V/I)² = R² + 1/(?C)² (V/I)² - R² = 1/(?C)² v[(V/I)² - R²] = 1/(?C)

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