Question

When the potential difference between the plates of a capacitor is halved, the magnitude of the electric energy stored in the capacitor U:

a. is 2U b. is U/2 c. remains the same d. is U/4

EXPLAIN.

Answer 1

The answer is option d. i.e

1. The energy stored in the capacitor is given by ;

Where: C is capacitance; V is the potential difference; U is the energy stored in the capacitor.

2. Now by substituting V/2 in the above formula, we see that the energy becomes 1/4th.

                                            (By the 1st equation).

As the energy is dependent on the square of the potential difference between the plates, the energy becomes 1/4th on reducing the potential difference to half of the original.

Getting into the equation:

The potential difference in the capacitor is given by

The small amount of work done in adding a small charge 'dq' to the capacitor is given by,

So, the total work done in adding a total charge of 'Q' to the capacitor is given by,

This integral becomes,       ; where C is a constant

Solving this,

We know that Q=CV. Substituting in the above equation we get,

Finally, this becomes,

This work done in filling the charges into a capacitor is stored as energy in it.

That is U=W(ideally)

So, the energy stored in the capacitor is

(I recommend you to go through the derivation once, even though just knowing the formula will do. If you have some idea how the formula comes it will be helpful in the future)