A certain substance has a dielectric constant of 2.8and a dielectric strength of 18 MV/m. If it is used as the dielectric material in a parallel plate capacitor, what minimum area should the plates of the capacitor have to obtain a capacitance of $\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{\mu F}$and to ensure that the capacitor will be able to withstand a potential difference of 4.0 kV?

• Dielectric constant of a substance, k = 2.8
• Dielectric strength of the substance, E =18MVm
• Capacitance of the plates,$C=7.0\times {10}^{-8}F$
• Potential difference,$V=4.0\times {10}^{3}V$

If the space between the plates of a capacitor is completely filled with a dielectric material, C the capacitance is increased by a factor k, called the dielectric constant, which is characteristic of the material. In a region that is completely filled by a dielectric, all electrostatic equations containing ${\mathbf{\in }}_0$must be modified by replacing ${\mathbf{\in }}_0$with role="math" localid="1661345220606" $\mathit{k}{\mathbf{\in }}_0$.

Using the relationship between electric field, potential difference, and distance, we can find the separation between the plates. Using the formula for the capacitance, we can find the area of the capacitor.

Formulae:

The capacitance of the plates due to dielectric material, $C=\frac{K·{\in }_0·A}{d}$ …(i)

Here, C is capacitance, A is area of cross section, d is separation between the two plates, k is dielectric constant of the material, ${\in }_0$is permittivity of the free space.

The electric strength of a field, $E=\frac{V}{d}$ …(ii)

Here, E is electric field, V is potential difference between the two plates, d is the separation between the two plates.

Dielectric strength is nothing but the electric field strength between parallel plate capacitor. Thus, using this value in equation (ii), we can get the value of the separation of the plates as given:

$$\begin{gathered} d=\frac{V}{E} \\ =\frac{4.0\times {10}^{3}V}{18\times {10}^{6}V/m} \\ =0.22\times {10}^{-3}m \end{gathered}$$

Now, suing this value in the equation (i), the minimum area of the plates can be calculated as follows:

$$\begin{gathered} A=\frac{Cd}{k{\in }_0} \\ =\frac{\left(7.0\times {10}^{8}F\right)·\left(0.22\times {10}^{-3}m\right)}{2.8·\left(8.85\times {10}^{-12}{C}^2/N·m^2\right)} \\ =0.63m^2 \end{gathered}$$

Hence, the value of the area is $0.63m^2$.