# Electric Potential

Electrostatic potential at any point of an electric field is defined as potential energy per unit charge at that point.

Electric Potential Energy:

Let us consider a system of two point charges $q$ and $q^{‘}$ (positive test charge) separated by a distance $r$

Figure:7.a

Charge $q$ is fixed at point $P$ and is displaced from point R to S along a radial line PRS shown in the figure.

According to Coulomb’s law, the magnitude of the force on a positive test charge as is given by,

If q’ moves towards $S$ through a small displacement $dr$ then work done by this force is given by,

Total work done from point $R$ to $S$ i.e., from $r_{1}$ to $r_{2}$ is given by,

It shows that work done on the test charge depends on end points not on the path taken.

The work done in moving the test charge $q’^{}$ from point $R$ to $S$ is equal to the change in potential energy in moving the test charge $q’^{}$ from point $R$ to $S$.

Thus,

where

is the potential energy of the test charge $q’^{}$ when it is at point $R$

is the potential energy of the test charge $q’^{}$ when it is at point $S$.

Thus potential energy of the test charge $q^{‘}$ at any distance r from charge $q$ is given by

## Electric potential:

Electric potential $V=U/q^{‘}$.

It is a scalar quantity and its SI unit is Volt.

1 Volt = 1 JC$^{-1}$

We have,

Now dividing both sides of the above equation by $q^{‘}$, we get,

Where $V_{r_{1}}$is the potential energy per unit charge at point $R$ and $V_{r_{2}}$is the potential energy per unit charge at point $S$.

If point $S$ is situated at infinity then,

(as Potential energy at infinity is zero)

Therefore, the electric potential at a point in an electric field is the ratio of work done in bringing a test charge from infinity to that point to the magnitude of the test charge.

## Potential due to point charge:

We have electric potential at a point due to a test charge $+q$,

Potential $V$ at any point due to all the point charges in a given system is given by,

For continuous charge distribution, the above equation becomes,

Let us consider a diagram showing the electric field $E$ due to a point charge $+q$ at point $O in a radially outward direction. Figure:7.b Consider a test charge$q^{‘}$is at point R. Now force on$q^{‘}$is,$F=q^{‘}E$Work done by the force in displacing$q^{‘}$from$R$to$S$is,$dW=F.dr=q^{‘}E.dr$Change in potential energy is,$dU=-dW=-q^{‘}E.dr$Change in electric potential is,$dV=dU/q^{‘}$Or,$dV=-E.dr$Or,$E=-\frac{dV}{dr}$The quantity$\frac{dV}{dr}\$ is the rate of change of potential with distance. This is known as potential gradient.