Question

5. A capacitor initially has a charge Q on it. As the charge is reduced what happens to the capacitance and the electric field between the plates?

   1. A) The capacitance and the electric field decrease.

   2. B) The capacitance is constant and the electric field decreases.

   3. C) The capacitance decreases and the electric field is constant.

   4. D) The capacitance and the electric field are both constant.

   5. E) There is not enough information to say.

Answer 1

capacitance

C = eo A / d

since capacitance is independent of charge, so it remains constant

electric field is given by

E = q/ ( eo A)

so electric field decreases with decrease in charge

so option (b) is correct

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