A parallel plate air filled capacitor has a capacitance of 50 pF.(a)If each of its plates has an area of$\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{}{\mathbf{m}}^{\mathbf{2}}$, what is the separation? (b)If the region between the plates is now filled with material having k = 5.6, what is the capacitance?

a) Capacitance with air filled capacitors,$C_0=50\times {10}^{-12}F$

b) Area of the plates,$A=0.35m^2$

c) Dielectric constant of the material,$k=5.6$

Using the formula of the capacitance of parallel plate capacitors we can find separation between the plates without dielectric and capacitance in presence of dielectric can be calculated by-product of the dielectric constant of the material and the capacitance of air-filled capacitors.

Formula:

The capacitance between the plates of a capacitor filled with a material,$C_0=\frac{k{\epsilon }_{0}A}{d}$…(i)

Where,

${\epsilon }_0=8.85\times {10}^{-12}\frac{C^2}{N.m^2}$is permittivity of free space

k = Dielectric constant of the material

d = Separation between the plates

A = Area of the plates

Using the given data, the separation between the plates can be calculated using equation (i) as follows: ( k = 1 )

$$\begin{gathered} d=\frac{{\epsilon }_{0}A}{C_0} \\ =\frac{\left(8.85\times {10}^{-12}{\displaystyle \frac{C^2}{N.m^2}}\right)\times \left(0.35m^2\right)}{\left(50\times {10}^{-12}F\right)} \\ =6.2\times {10}^{2}m \end{gathered}$$

Hence, the value of the separation is$6.2\times {10}^{2}m$ .

According to the capacitance equation (i), we can get the capacitance value with dielectric material filled as follows:

$$\begin{gathered} {\mathrm{C}}_{\mathrm{K}}=\mathrm{k}\times {\mathrm{C}}_0 \\ =5.6\times \left(50\times {10}^{-12}\mathrm{F}\right) \\ =2.8\times {10}^2\mathrm{pF} \end{gathered}$$

Hence, the value of the capacitance is$2.8\times {10}^2\mathrm{pF}$ .