Question

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?

Short answer

a) The separation between the plates is$6.2\times {10}^{-2}m$ .

b) The capacitance between the capacitor plates is$2.8\times {10}^2\mathrm{pF}$ .

Step-by-step solution

The given data

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,$k=5.6$

Understanding the concept of the capacitance

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

(a) Calculation of the separation between 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$6.2\times {10}^{2}m$ .

(b) Calculation of the capacitance value

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$2.8\times {10}^2\mathrm{pF}$ .