The capacitance of a variable capacitor joined with the battery of $100$V is changed from $2\mu F$ to $10\mu F$. What is the change in the energy stored in it?

Given:

$Voltage=V=100V$

$C_1=2\mu F$ (Initial capacitance)

$C_2=10\mu F$ (Final capacitance)

As $E=\frac{1}{2}C{V}^2$

Now

$E_1=\frac{1}{2}{C}_1{V}^2=\frac{1}{2}\times 2\times {10}^{-6}\times V^2$

$E_2=\frac{1}{2}{C}_2{V}^2=\frac{1}{2}\times 10\times {10}^{-6}\times V^2$

$\Delta E=E_2-E_1=\frac{1}{2}\times 10\times {10}^{-6}\times V^2-\frac{1}{2}\times 2\times {10}^{-6}\times V^2$

$\Delta E=\frac{1}{2}\times 8\times {10}^{-6}\times V^2$

$\Delta E=\frac{1}{2}\times 8\times {10}^{-6}\times {100}^2$

$\Delta E=4\times 10^{-2}J$

Hence the correct option is (D).