# Capacitance of parallel plate capacitor with dielectric medium

We have discussed the basic things of a capacitor in previous article. That article covered the definition, construction, types, formula and uses of a capacitor. In this article, we are going to discuss capacitance of parallel plate capacitor and its behavior in presence of dielectric medium or slab between the plates of the capacitor.

1. How to construct a Parallel plate capacitor?
2. Formula for capacitance of parallel plate capacitor.
3. Derivation of Capacitance of a parallel plate capacitor.
4. On which parameters does the capacitance of parallel plate capacitor depend?
5. Capacitance of parallel plate capacitor with dielectric medium.
6. Uses of parallel plate capacitor.

## How to construct a Parallel plate capacitor?

A parallel plate capacitor is a system of equally and oppositely charged two conductors placed at some distance of separation. To construct a parallel plate capacitor we need to place two conducting plates at a small separation. The plates should be equally and oppositely charged. Shape of the plates may be rectangular or circular.

## Formula for capacitance of parallel plate capacitor

The general formula for any type of capacitor is, Q=CV, where Q is the electric charge on each plate, V is the potential across the plates and C is the capacitance of the capacitor.

Now, a parallel plate capacitor has a special formula for its capacitance. If the cross-section area of each plate be A and the distance between the plates be d, then the formula for capacitance of the parallel plate capacitor is, $\small {\color{Blue} C=\frac{\epsilon _{0}A}{d}}$

If the space between the plates is filled with a dielectric medium of dielectric constant K then from the above formula, $\small {\color{Blue} C=\frac{K \epsilon _{0}A}{d}}$

## Derivation of Capacitance of a parallel plate capacitor

Let us consider a parallel plate capacitor with two plates of cross-section area A, separated by a distance d. The medium between the plates is air medium. If V be the voltage applied across the plates of the capacitor, then the electric field inside the capacitor will be, $\small {\color{Blue} E=\frac{V}{d}}$ ……………..(1)

Again, from the Gauss’s law of electrostatics one can get the electric field inside a capacitor of two oppositely charged plates is, $\small {\color{Blue} E=\frac{\sigma }{\epsilon _{0}}}$ …………….(2)

Where, $\small {\color{Blue} \sigma }$ is the surface charge density of each plates.

From equation-1 and equation-2 we get, $\small {\color{Blue} \frac{V}{d}}={\color{Blue} \frac{\sigma }{\epsilon _{0}}}$

or, $\small {\color{Blue} V}={\color{Blue} \frac{\sigma d}{\epsilon _{0}}}$

or, $\small {\color{Blue} V}={\color{Blue} \frac{Q d}{\epsilon _{0} A}}$

Since, the surface charge density, $\small \sigma =\frac{Q}{A}$

Now, the capacitance of the parallel plate capacitor is, $\small {\color{Blue} C=\frac{Q}{V}}$

or, $\small {\color{Blue} C=\frac{\epsilon _{0}A}{d}}$ ……………………(3).

This is the formula for the capacitance of a parallel plate capacitor.

Now, if a dielectric medium of dielectric constant K is inserted in the region between the plates then, $\small {\color{Blue} C=\frac{K \epsilon _{0}A}{d}}$.

## On which parameters does the capacitance of parallel plate capacitor depend?

From equation-3 one see that the capacitance of a parallel plate capacitor depends on the following parameters –

1. The cross-section area of the plates. A Greater cross-section area gives greater value of capacitance.
2. Distance between the plates. Capacitance is inversely proportional to the distance between the plates.
3. The medium between the plates. The presence of dielectric medium between the plates increases then capacitance of the capacitor.

## Capacitance of parallel plate capacitor with dielectric medium

Let a parallel plate capacitor with plate area A and the distance between the plates d.

When the medium between the plates is air medium, the capacitance, $\small {\color{Blue} C=\frac{\epsilon _{0}A}{d}}$.

And, when a dielectric slab of dielectric constant K is inserted between the plates, the capacitance, $\small {\color{Blue} C=\frac{K\epsilon _{0}A}{d}}$.

So, the capacitance of a parallel plate capacitor increases due to inserting a dielectric slab or dielectric medium between the plates of the capacitor. New value of the capacitance becomes K times the capacitance when the medium between the plates is air.

## Effects on capacitance, charge and voltage due to inserting the dielectric slab

1. Due to inserting of dielectric slab the capacitance of the capacitor increases.
2. If the capacitor is in charging mode i.e. if the battery is on, then voltage across the capacitor remains same and hence the amount of charge on each plate increases. But if the battery is off, then the charge on the plates remains same and the voltage across the plates decreases to maintain Q=CV formula.

## Uses of a capacitor

Capacitors have huge applications in our daily life. In ceiling fan, TV and other electric devices use capacitor for different purpose. A capacitor can be used in filter circuit. Overall, capacitor is one of the most useful circuit components.

Homework Problems:

1. A parallel plate capacitor has d=2 mm and A=5 cm2. If the medium between the plates be air medium and 10 volt voltage is applied across the capacitor then find the electric charge of the capacitor. Now, if you insert a dielectric slab (K=2) then what will be the charge of the capacitor?
2. Suppose you have inserted a dielectric slab (K=3) between the plates of a parallel plate capacitor of capacitance C. What will be the ratio of new capacitance to the previous capacitance of the capacitor?

This is all from the parallel plate capacitor. If you have any doubt on this topic you can ask me in the comment section.

Thank you!

Related posts:

1. Capacitance of different type of capacitors – Spherical, Cylindrical, parallel plate and isolated conductors.
2. Energy stored in a capacitor
3. Capacitance of Earth and other planets.