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 , shown in the first pic. The capacitor (C0=e0A/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.

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

Explanation / Answer

Capacitance is proportional to Area/distance (A) We also know Q = V*C (B) 1) from (A) we know capacitance of capacitor2 is twice that of capacitor 1, and from (B) we know voltage is inversly proportional to capacitance so voltage across capacitor2 is V/2 2) from (A) we want 2A/x to be equal to A/d thus x, the distance between the plates on capacitor2 must be 2d (so that the 2s cancel giving A/d = A/d)

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