Question 187: Electrostatics & Capacitance

Question

A parallel plate capacitor has capacitance $C$, when there is vacuum within the parallel plates. A sheet having thickness $\left(\frac{1}{3}\right)^{\mathrm{rd}}$ of the separation between the plates and relative permittivity $K$ is introduced between the plates. The new capacitance of the system is :

Accepted Answer

For a parallel plate capacitor with plate area $A$, separation $d$, and vacuum between the plates, the capacitance $C$ is given by:
$$ C=\frac{\epsilon_0 A}{d} $$
A dielectric sheet of thickness $t=\frac{d}{3}$ and relative permittivity (dielectric constant) $K$ is inserted.
So, the system is made of two capacitors connected in series:
Capacitor $1\left(\mathrm{C}_1\right)$ : A capacitor filled with the dielectric material of thickness $\mathrm{t}=\frac{\mathrm{d}}{3}$.
Capacitor $2\left(\mathrm{C}_2\right)$ : A capacitor with vacuum filling the remaining space of thickness $(\mathrm{d}-\mathrm{t})=\mathrm{d}- \frac{\mathrm{d}}{3}=\frac{2 \mathrm{~d}}{3}$.
For the capacitor $\left(\mathrm{C}_1\right)$ :
$$ C_1=\frac{K \epsilon_0 A}{t}=\frac{K \epsilon_0 A}{d / 3}=\frac{3 K \epsilon_0 A}{d}=3 K C $$
For the capacitor $\left(\mathrm{C}_2\right)$ :
$$ C_2=\frac{\epsilon_0 A}{d-t}=\frac{\epsilon_0 A}{2 d / 3}=\frac{3 \epsilon_0 A}{2 d}=\frac{3}{2} C $$
Since they are in series, the equivalent capacitance is :
$$ \frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2} $$
Substituting the values :
$$ \frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{3 \mathrm{KC}}+\frac{1}{3 / 2 \mathrm{C}} $$
$$\Rightarrow $$ $\frac{1}{C_{e q}}=\frac{1}{3 K C}+\frac{2}{3 C}$
$$\Rightarrow $$ $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1+2 \mathrm{~K}}{3 \mathrm{KC}}$
$$\Rightarrow $$ $$ C_{\mathrm{eq}}=\frac{3 K C}{2 K+1} $$
Therefore, the new capacitance of the system is $\frac{3 \mathrm{KC}}{2 \mathrm{~K}+1}$.
Hence, the correct option is (D).

Answer

$\frac{3 C K^2}{(2 K+1)^2}$

Answer

$\frac{4 K C}{3 K-1}$

Answer

$\frac{C K}{2+K}$

Answer

$\frac{3 K C}{2 K+1}$

Electrostatics & Capacitance

A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness \left(\frac{1}{3}\right)^{\mathrm{rd}} of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :

Choose the correct answer: