Aparallel-plate capacitor is connected to a battery. what happens to the stored energy if the plate separation is doubled while the capacitor remains connected to the battery? (a) it remains the same(b) it is doubles(c) it decreases by a factor of 2(d) it decreases by a factor of 4(e) it increases by a factor of 4

option (b)

Explanation:

The capacitance of a parallel plate capacitor is given by

Where, A be the area of plates, d be the distance between the plates

Let the potential difference of the battery is V.

the energy stored in the capacitor is given by

Substitute the value of C, we get

  .... (1)

Now the distance between the plates is doubled, so the new capacitance

the new energy stored in the capacitor is given by

(As battery remains connected so the potential difference remains constant)

So, the energy is doubled.