Easy One Out

Well guys, here is the Answer for you : U = $\frac{1}{2}\times(C \times V^{2})$ = $\frac{1}{2} \times \frac{ k \times A \times V^{2}}{d}$

Here, k = Permittivity constant of free space = $8.85 \times 10^{-12}$ $C^{-1}$ $N^{2}$ $m^{-2}$ U = Energy stored in the capacitor.

When the separation between the plates is decreased by 10%, the energy stored becomes U* such that,

$\frac{U* - U}{U} \times 100$ = $\frac{10}{9} - 1 \times 100$ = $\frac{100}{9}$ => $11.11\%$

We can use a variety of methods. I used plugging in a dummy value.

If Voltage = 10 and separation = 10, then energy = (100 / 10) = 10 Joules

Now, if we decrease the separation by 10%, thus making it 9, we have energy = (100 / 9) = 11.111... J

So, percentage change = ((11.11 - 10) / 10) x 100 = 11.11% is the answer