Let’s consider a **parallel plate capacitor** having plate area A and equal and opposite **charges** q on both of the plates.

The seperation distance between the plates of the capacitor is d which is much smaller than the plate **surface area** A.

The space between the plates is filled with the material of **dielectric constant** K and the **permittivity of the material medium** is ε. See the diagram below.

## DERIVATION PROCESS

[latexpage]

First of all let’s find the relationship between the **relative permittivity of the medium**$\epsilon_r$ and the **permittivity of the medium** $\epsilon$. The relationship is given as :

$$

\epsilon_r=K=\frac{\epsilon}{\epsilon_0}

$$

From this equation we can write the formula of permittivity of the medium in the form of relative permittivity of the medium also called **dielectric constant** and the **absolute permittivity of the free space.** The formula is given as-

$$

\quicklatex{size=20}

\epsilon=\epsilon_r\epsilon_0=K\epsilon_0

$$

Now the **electric field** between the plates is given as-

\begin{align*}

E&=\frac{\sigma}{\epsilon}\\

&=\frac{\sigma}{K\epsilon_0}\\

\end{align*}

Where the value of $\sigma$ is charge per unit surface area, that is-

$$\sigma=\frac{q}{A}$$ put it in the formula of **electric field**, which we have obtained above –

$$

E=\frac{q/A}{K\epsilon_0}=\frac{q}{\epsilon_0KA}

$$

Now the relationship between the **electric potential difference** V and the electric field E is V=Ed. Now after substituting the value of electric field in the formula of electric potential, we get –

$$

V=\frac{qd}{\epsilon_0KA}

$$

Now the **capacitance** C is given as charge(q) per unit **potential difference** (V).

\begin{align*}

\quicklatex{size=20}

C&=\frac{q}{V}\\

&=\frac{q}{\left(\frac{qd}{\epsilon_0KA}\right)}\\

C&=\frac{\epsilon_0KA}{d}

\end{align*}

This is the required value of capacitance of a **parallel plate capacitor**, when a dielectric material is inserted between the plates.

## FREQUENTLY ASKED QUESTIONS

1). **What is the formula of capacitance of parallel plate capacitor?**

**Ans**– The formula of capacitance of parallel plate capacitor when a dielectric material is inserted between the plates is $$C=\frac{\epsilon_0KA}{d}$$ and when vaccum is filled between the plates then the capacitance is – $$C=\frac{\epsilon_0A}{d}$$

2). **What is capacitor derive an expression for capacitance of parallel plate capacitor?**

**Ans**– Capacitor is a passive device which is used to stored electrical charges in electric field.

Go through this article for complete derivation. **Expression for capacitance of parallel plate capacitor**.

3). **What do you mean by parallel plate capacitor?**

Ans– A parallel plate capacitor is an arrangement of two metal plates connected in parallel separated from each other by some distance. A **dielectric** material is filled between the gap of the plates. … This dielectric material does not allow the flow of electric current through it, due to its **non-conductive** property.