91Ó°ÊÓ

Capacitance is the measure or the capacity of an electric device to store charge and allow the device to use the charge up as desired.The difference in potential causes the charge to get stored on the plates of the conductor.

Mathematically,

$\mathbf{C}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{V}}$

Where Q is the total charge stored and V is the potential difference or the voltage applied electronically.

The given data can be listed below,

• The dielectric constant is, K= 3.6
• The dielectric strength is, $E=1.6×{10}^{7}V/m$
• The capacitor is, $C=1.25×{10}^{-9}\text{F}$
• The potential difference is,V = 5500 V .
• The minimum area of the plates is, A

The expression of capacitance for a parallel plate capacitor with dielectric is,

$C=\frac{K{e}_{0}A}{d}$ ...(i)

Here, K is the dielectric constant, is the permittivity, A is the area of plates and d is the distance between the plates.

Solve for Area,

$A=\frac{Cd}{K{e}_0}$ ...(ii)

Now, the potential applied is given as,

V = Ed

Here, E is the electric field magnitude. Solve for d,

$d=\frac{V}{E}$

Substitute the 5500 V for V, and for E in the above equation.

$$\begin{gathered} =\frac{\text{5500 V}}{\text{1.60}×{\text{10}}^7\text{V/m}} \\ =\text{3.44}×{\text{10}}^{-4}\text{m.} \end{gathered}$$

Substitute all the values in equation (ii),

$$\begin{gathered} A=\frac{\text{(1.25}×{\text{10}}^{-9}\text{F}\left)\right(3.44×{10}^{-4}\text{m})}{\left(360\right)(8.854×{10}^{-12}{C}^{2}N×m^2} \\ =0.0135{\text{m}}^{\text{2}} \end{gathered}$$

Thus, the area of the plates is$0.0135{\text{m}}^{\text{2}}$.