The charge on the square plates of a parallel-plate capacitor is Q. The potential across the plates is maintained with constant voltage by a battery as the two plates of the capacitor are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. (i) The amount of charge on the plates is now equal to [3] A) 4Q. B) 2Q. C) Q. D) Q/2. E) Q/4. (ii) What happens to the amount of charge on the plates if the plates are not connected to the battery when they are pulled apart?[2]

Question

Stephenson

Physics

2 February, 13:37

The charge on the square plates of a parallel-plate capacitor is Q. The potential across the plates is maintained with constant voltage by a battery as the two plates of the capacitor are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. (i) The amount of charge on the plates is now equal to [3] A) 4Q. B) 2Q. C) Q. D) Q/2. E) Q/4. (ii) What happens to the amount of charge on the plates if the plates are not connected to the battery when they are pulled apart?[2]

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Answers (1)

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Fatima Blake · Answer 1

i) Q' = Q/2.

ii) If the wires to the battery are disconnected, the amount of charge on the plates remains constant, and the voltage across the plates remains the same.

Explanation:

i) The formula for capacitance is given as

C = A/d

Where d = separation between the plates

The formula for the charge stored is given as

Q = CV

Q = AV/d

When "d" is doubled, the charge stored "Q" becomes half because Q is inversely proportional to the separation between the plates, d.

Thus, Q' = Q/2.

ii) If the wires to the battery are disconnected, the amount of charge on the plates remains constant, and the voltage across the plates remains the same.