#
Derive expression for energy stored in a parallel plate capacitor of capacitance C with air as medium between its plates having charges Q & -Q.Show that this energy can be expressed in terms of electric field as 1/2e_{0}E^{2}Ad where A is the area of each plate and d is the seperation between each plate.How will the energy stored in a fully charged capacitor change when the seperationbetween the plates is doubled and a dielectric medium of dielectric constant 4 is introduced between the plates ?

**Energy Stored in a Charged Capacitor **

The energy of a charged capacitor is measured by the total work done in charging the capacitor to a given potential.

Let us assume that initially both the plates are uncharged. Now, we have to repeatedly remove small positive charges from one plate and transfer them to the other plate.

Let

*q* →Total quantity of charge transferred

*V* →Potential difference between the two plates

Then,

*q* = *CV*

Now, when an additional small charge *dq*is transferred from the negative plate to the positive plate, the small work done is given by,

$dw=\left(\frac{Q}{C}\right).dq$

The total work done in transferring charge *Q*is given by,

or,** $U=\int dw=\int \frac{\{Q.dq\}}{C}=\frac{{Q}^{2}}{2C}=\frac{1}{2}C{V}^{2}$**

Refer the following link for the derivation of energy density:

https://www.meritnation.com/ask-answer/question/what-is-energy-density-of-an-electric-field-in-capicators/electrostatic-potential-and-capacitance/1096595

Let initially the capacitance is, $C=\frac{{\in}_{0}A}{d}$

When the separation between the plates is doubled and a dielectric medium of dielectric constant 4 is introduced between the plates then the capacitance is given by:

${C}_{1}=\frac{4{\in}_{0}A}{2d}=2\frac{{\in}_{0}A}{d}=2C$

Hope this answers your question. In case of further query feel free to ask from our experts.

**
**