Answer
(a) $U_o = 3.6 \times 10^{-3} \,\text{J}$ and $U_K = 13.5 \times 10^{-3} \,\text{J}$
(b) $\Delta U = 9.9 \times 10^{-3} \,\text{J}$ increase.
Work Step by Step
(a) The energy storage before the dielectric material is given by
\begin{align*}
U_o &= \dfrac{1}{2} CV^2\\
& = \dfrac{1}{2} (12.5 \times 10^{-6} \,\text{C}) (24\,\text{V})^2 \\
&=\boxed{ 3.6 \times 10^{-3} \,\text{J}}
\end{align*}
After the dielectric material with a dielectric constant is added $K = 3.75$ the energy stored will be given by
\begin{align*}
U_K &= \dfrac{1}{2}K CV^2\\
& = \dfrac{1}{2} (3.75 )(12.5 \times 10^{-6} \,\text{C}) (24\,\text{V})^2 \\
&=\boxed{ 13.5 \times 10^{-3} \,\text{J}}
\end{align*}
(d) The energy will change by
$$\Delta U = U_K -U_o = 13.5 \times 10^{-3} \,\text{J}- 3.6 \times 10^{-3} \,\text{J} = +9.9 \times 10^{-3} \,\text{J}$$
The energy stored is increased.
