Answer
See the detailed answer below.
Work Step by Step
a)
The electric field between the capacitor's parallel plates is given by
$$E=\dfrac{V}{d}=\dfrac{35}{3\times10^{-3}}=\color{red}{\bf 1.167\times 10^{4}}\;\rm V/m$$
The capacitance of the described capacitor is given by
$$C=\dfrac{\epsilon_0A}{d}=\dfrac{8.85\times10^{-12}\cdot 2 }{3\times10^{-3}}=\color{red}{\bf 5.9\times 10^{-9}}\;\rm F$$
Now we can find the charge of the capacitor which is given by
$$Q=CV=5.9\times 10^{-9}\cdot 35=\color{red}{\bf 2.065\times 10^{-7}}\;\rm C$$
The energy stored in the capacitor is given by
$$PE=\frac{1}{2}CV^2=\frac{1}{2}\cdot5.9\times 10^{-9}\cdot 35^2= \color{red}{\bf 3.61\times 10^{-6}}\;\rm J$$
b)
We need to repeat the same answer above by considering the dielectric materials added to the capacitor between its two plates.
The electric field between the capacitor's parallel plates is given by
$$E=\dfrac{V}{d}=\dfrac{35}{3\times10^{-3}}=\color{red}{\bf 1.167\times 10^{4}}\;\rm V/m$$
The new capacitance of the described capacitor is given by
$$C=\dfrac{\epsilon_0AK}{d}=\dfrac{8.85\times10^{-12}\cdot 2 \cdot 3.2}{3\times10^{-3}}=\color{red}{\bf 1.89\times 10^{-8}}\;\rm F$$
The new charge of the capacitor
$$Q=CV=1.89\times 10^{-8}\cdot 35=\color{red}{\bf 6.61\times 10^{-7}}\;\rm C$$
The new energy stored in the capacitor is
$$PE=\frac{1}{2}CV^2=\frac{1}{2}\cdot 1.89\times 10^{-8}\cdot 35^2= \color{red}{\bf 1.156\times 10^{-5}}\;\rm J$$
