# Parallel Plate Capacitor

A parallel plate capacitor has two parallel plates, each of area $A$, separated by a distance $d$. One plate is given a positive charge and other an equal negative charge. The capacitance of a parallel plate capacitor is given by \begin{align} C=\frac{\epsilon_0 A}{d}. \end{align}

If interior of a parallel plate capacitor is filled with a dielectric material of dielectrinc constant $k$ then its capacitance is given by \begin{align} C=\frac{\epsilon_0 k A}{d}. \end{align}

The electric field in a parallel plate capacitor is constant.

The force between plates of a parallel plate capacitor is given by \begin{align} F=\frac{Q^2}{2A\epsilon_0} \end{align}

## Problems from IIT JEE

Problem (IIT JEE 2015): A parallel plate capacitor having plates of area $s$ and plate separation $d$, has capacitance $C_1$ in air. When two dielectrics of different relative permittivities ($\epsilon_1=2$ and $\epsilon_2=4$) are introduced between the two plates as shown in the figure, the capacitance becomes $C_2$. The ratio $C_2/C_1$ is,

1. 6/5
2. 5/3
3. 7/5
4. 7/3

Solution: The capacitance of an air filled parallel plate capacitor having plates of area $s$ and plate separation $d$ is given by $C_1=\epsilon_0 s/d$. Consider the case when two dielectrics of relative permittivities ($\epsilon_1=2$ and $\epsilon_2=4$) are introduced between the two plates as shown in the figure. We can think of this capacitor to be made up of three parts:

1. Capacitor of plate area $s/2$, plate separation $d$, and filled with a dielectric of relative permittivity $\epsilon_1=2$ (lower half). The capacitance of this part is $C_a=\epsilon_1 \epsilon_0\frac{ s/2}{d}=\epsilon_0 \frac{s}{d}=C_1$.
2. Capacitor of plate area $s/2$, plate separation $d/2$, and filled with a dielectric of relative permittivity $\epsilon_1=2$ (left upper half). The capacitance of this part is $C_b=\epsilon_1 \epsilon_0 \frac{s/2}{d/2}=2\epsilon_0 \frac{s}{d}=2C_1$.
3. Capacitor of plate area $s/2$, plate separation $d/2$, and filled with a dielectric of relative permittivity $\epsilon_2=4$ (right upper half). The capacitance of this part is $C_c=\epsilon_2 \epsilon_0 \frac{s/2}{d/2}=4\epsilon_0 \frac{s}{d}=4C_1$.
The capacitors $C_b$ and $C_c$ are in series with equivalent capacitance, \begin{align} C_{bc} & =\frac{C_b C_c}{C_b+C_c} \\ &=\frac{(2C_1)(4C_1)}{(2C_1)+(4C_1)} \\ &=\frac{4}{3}C_1.\nonumber \end{align} The capacitor $C_{bc}$ is in parallel with $C_a$. The equivalent capacitance is, \begin{align} C_2 &=C_{abc} \\ &=C_{a}\parallel C_{bc} \\ &=C_1+\frac{4}{3}C_1 \\ &=\frac{7}{3}C_1.\nonumber \end{align}

## Exercise

1. You prepare a capacitor by placing aluminium foils on either side of an A4 sheet of paper. The A4 size is 8.5 inch x 11 inch and a pack of 500 such sheets measure 5 cm in thickness. How much is the capacitance?