The plates of a parallel-plate capacitor are 3.28 mm apart, and each has an area of $\mathbf{9}\mathbf{.}\mathbf{82}\mathit{c}{\mathit{m}}^{\mathbf{2}}$. Each plate carries a charge of magnitude $\mathbf{4}\mathbf{.}\mathbf{35}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{8}}$C. The plates are in vacuum. What is (a) the capacitance; (b) the potential difference between the plates; (c) the magnitude of the electric field between the plates?

Capacitance:

$\mathit{C}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\frac{\mathbf{A}}{\mathbf{d}}$ (1)

Where d is the distance between the plates andis the area of each plate

Potential difference:

${\mathit{V}}_{\mathbf{a}\mathbf{b}}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{C}}$ (2)

Where Q is the charge on conducting plates.

Electric field:

$\mathit{E}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{a}\mathbf{b}}}{\mathbf{d}}$ (3)

Due to the charge on the two parallel plates, there is electrical energy stored between the two plates called the capacitance, where the capacitance depends on the area of the plates, and it is given by equation (1).

So, it can be calculated as

$C=(8.854*{10}^{-12})\left(\frac{9.82*{10}^{-4}}{0.00328}\right)/)=2.65*{10}^{-12}F=2.65pF$

The capacitance depends on the potential difference between the two plates as given by equation (2).

It can be calculated as

$V_{ab}=\frac{4.35*{10}^{-8}}{2.65*{10}^{-12}}=16.40kV$

Due to the charged particles, there is a uniform electric field between them and is given by equation (3). It can be calculated as

$E=\frac{16400}{0.00328}=5*{10}^{6}V/m$

Hence, the capacitance is 2.65pF. The potential difference between the plates is . The magnitude of the electric field between the plates is $5*{10}^{6}V/m$.