Well, In this article, we are going to derive the expression for the capacitance of a parallel plate capacitor when a dielectric slab of thickness t is inserted between the plates. Wants to know all about the derivation, stay tuned with us till the end…

# DERIVATION FOR THE CAPACITANCE

Let’s take a **parallel plate capacitor** having a plate surface area is **A**. A dielectric slab of thickness t is inserted between the plates. And the dielectric slab has the **dielectric constant** K and the permittivity of its medium is ε. See the figure below :

After putting the **dielectric** slab of thickness t the remaining distance will be (d – t). Let the electric field inside the dielectric slab is E and the field inside the plates and outside the slab is $E_0$.

## DERIVATION PROCESS

The **electric field** inside the slab is given as-

$$

E=\frac{\sigma}{\epsilon}=\frac{\sigma}{K\epsilon_0}=\frac{q}{\epsilon_0KA}

$$

And the **electric field** outside the slab is given as –

$$

E_0=\frac{\sigma}{\epsilon_0}=\frac{q}{\epsilon_0A}

$$

The total **potential difference** between the plates is given as the sum of the **potential difference** due to the slab and without the slab. Which is clearly written below :

\begin{align*}

V&=E_0\left(d-t\right)+Et\\

&=\frac{q\left(d-t\right)}{\epsilon_0A}+\frac{qt}{\epsilon_0KA}\\

&=\frac{Kq\left(d-t\right)+qt}{\epsilon_0KA}\\

&=\frac{Kqd-kqt+qt}{\epsilon_0KA}\\

&=\frac{q\left[kd-kt+t\right]}{\epsilon_0KA}

\end{align*}

**Capacitance of any capacitor is given as charge per unit potential difference**. Let’s put the value of potential difference in the formula of capacitance.

\begin{align*}

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

&=\frac{q}{\left(\frac{q\left[kd-kt+t\right]}{\epsilon_0KA}\right)}\\

C&=\frac{\epsilon_0KA}{Kd-kt+t}\\

&=\frac{\epsilon_0KA}{Kd+t\left(1-K\right)}

\end{align*}

This is the required value of capacitance of a parallel plate capacitor when a dielectric slab of thickness t is inserted between the plates.

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