# 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,

- 6/5
- 5/3
- 7/5
- 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:

- 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 $C_a=\epsilon_1 \epsilon_0\frac{ s/2}{d}=\epsilon_0 \frac{s}{d}=C_1$.
- 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 $C_b=\epsilon_1 \epsilon_0 \frac{s/2}{d/2}=2\epsilon_0 \frac{s}{d}=2C_1$.
- 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

- 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?

## Related

- Capacitance
- Capacitors in Series and Parallel
- Capacitors from kitchen utensils
- Energy Stored in Capacitor
- Charging and discharging of a capacitor
- Electric field in a capacitor