Question 72: Electrostatics & Capacitance

Question

A parallel plate capacitor is having separation between plates 0.885 mm . It has a capacitance of $1 \mu \mathrm{~F}$ when the space between the plates is filled with an insulating material of resistivity $1 	imes 10^{13} \Omega \mathrm{~m}$ and resistance $17.7 	imes 10^{14} \Omega$. Relative permittivity of the insulating material is $\alpha 	imes 10^7$. The value of $\alpha$ is $\_\_\_\_$ .
(Take permittivity of free space $=8.85 	imes 10^{-12} \mathrm{~F} / \mathrm{m}$ )

Accepted Answer

The resistance of a conductor of length d and cross-sectional area A is :
$$ \mathrm{R}=\frac{ho \mathrm{d}}{\mathrm{~A}} $$
Where, $ho$ is the resistivity of the material.
The capacitance of a parallel plate capacitor with plate separation d and area A is :
$$ C=\frac{\epsilon A}{d}=\frac{\epsilon_{\mathrm{r}} \epsilon_0 A}{d} $$
$$\Rightarrow $$ $\mathrm{R} 	imes \mathrm{C}=\left(\frac{ho \mathrm{d}}{\mathrm{A}}ight) 	imes\left(\frac{\epsilon_{\mathrm{r}} \epsilon_0 \mathrm{~A}}{\mathrm{~d}}ight)$
$$\Rightarrow $$ $\mathrm{RC}=ho \epsilon_{\mathrm{r}} \epsilon_0$
$$\Rightarrow $$ $$ \epsilon_{\mathrm{r}}=\frac{\mathrm{RC}}{ho \epsilon_0} $$
The resistance of material is $\mathrm{R}=17.7 	imes 10^{14} \Omega$.
The capacitance is $\mathrm{C}=1 \mu \mathrm{~F}=1 	imes 10^{-6} \mathrm{~F}$
The resistivity is $ho=1 	imes 10^{13} \Omega \mathrm{~m}$
And the permittivity of free space is $\epsilon_0=8.85 	imes 10^{-12} \mathrm{~F} / \mathrm{m}$
Putting the values for relative permittivity,
$$ \epsilon_{\mathrm{r}}=\frac{\left(17.7 	imes 10^{14}ight) 	imes\left(1 	imes 10^{-6}ight)}{\left(1 	imes 10^{13}ight) 	imes\left(8.85 	imes 10^{-12}ight)} $$
$$\Rightarrow $$ $\epsilon_{\mathrm{r}}=\frac{17.7 	imes 10^8}{8.85 	imes 10^1}$
$$\Rightarrow $$ $$ \epsilon_{\mathrm{r}}=2 	imes 10^7=\alpha 	imes 10^7 \Rightarrow \alpha=2 $$
Therefore, the value of $\alpha$ is 2 .

Electrostatics & Capacitance

Integer / Numeric Answer Type