__What is Electric Potential Energy?__

*Electric potential **energy* *at a point in the **electric field** produced by a **charge** is defined as the **work* *done by an external **force** in bringing a test charge from infinity to that point.*

Consider that a positive test charge ( + q_0 ) is placed in an electric field ( \vec {E} ) produced by another positive source charge ( + Q ) as shown in figure. The *electrostatic force* of repulsion acting on the test charge will be –

\vec {F_e} = q_0 \vec {E}

This force has a tendency to move the test charge in the direction of the electric field. Now, suppose that an external force ( \vec {F_0} ) is acting on the test charge just to resist the electrostatic force ( \vec {F_e} ) . So, the test charge slowly moves without any *acceleration* towards the source charge ( Q ) .

Therefore, \quad \vec {F_0} = - \vec {F_e} = - q_0 \vec {E} …….. (1)

Work done by the external force in displacing the test charge through a small distance ( d \vec {r} ) from point A to point B in the electric field, will be –

W_{AB} = \int\limits_{B}^{A} \vec {F_0} . d \vec {r}

= - \int\limits_{B}^{A} q_0 \vec {E} \ d \vec {r}

= - q_0 \int\limits_{B}^{A} \vec {E} . d \vec {r} . …… (2)

*This work done gives the difference in electrical potential energy between two points A and B within the electric field.*

Therefore, \quad W_{AB} = ( U_B - U_A )

Hence, \quad ( U_B - U_A ) = - q_0 \int\limits_{B}^{A} \vec {E} . d \vec {r}

= - q_0 \int\limits_{B}^{A} \vec {E} . d \vec {r} ……… (3)

### Potential Energy at infinity

Consider that, the point A is placed at infinity from charge ( + Q ). Then, there will be no *electrostatic force* between charge ( + Q ) and test charge ( + q_0 ) .

Thus potential energy of the test charge at infinity is zero. Thus, work done in moving the test charge from infinity to the point B will be –

( U_B - U_{\infty} ) = W_{\infty B}

Or, \quad ( U_B - 0 ) = W_{\infty B}

So, \quad U_B = W_{\infty B}

Therefore, equation (3) reduces to the form –

U_B = W_{\infty B} = - q_0 \int\limits_{B}^{\infty} \vec {E} . d \vec {r}

= - q_0 \int\limits_{B}^{\infty} \vec {E} . d \vec {r} ………. (4)

This gives the electric potential energy at point ( B ) .

__Electric Potential__

*Electric potential** at a point in an electric field is defined as the electric potential energy per unit charge at that point.*

Let, ( U_B ) is the electric potential energy at a point B . Then electric potential at that point is given by –

V = \left ( \frac {U_B}{q_0} \right ) …….. (5)

By equation (4), the potential energy at point B will be –

U_B = - q_0 \int\limits_{B}^{\infty} \vec {E} . d \vec {r} ……. (4)

Therefore, electric potential at point B will be –

V_B = \left ( \frac {U_B}{q_0} \right ) = - \int\limits_{B}^{\infty} \vec {E} . d \vec {r}

= - \int\limits_{B}^{\infty} \vec {E} . d \vec {r} ………. (6)

Electric potential is a scalar quantity. In SI system unit of electric potential is **volt (V)**.

Since, \quad V = \left ( \frac {U}{q_0} \right )

Therefore, \quad 1 \ \text {Volt} = \frac {1 \ \text {joule ( J )}}{1 \ \text {coulomb ( C )}} = 1 \ \text {J - C}^{-1}

__Potential difference__

*Potential difference between two points in an electric field is defined as the electric potential energy difference per unit charge.*

Let, ( \Delta U ) is the electric potential energy difference between two points ( A ) \ \text {and} \ ( B ) in an *electric field*. Therefore –

\Delta U = ( U_B - U_A )

Then, electric potential difference ( \Delta V ) between the points A and B will be –

\Delta V = ( V_B - V_A ) = \left ( \frac {\Delta U}{q_0} \right )

Or, \quad \Delta V = \frac {( U_B - U_A )}{q_0}

So, \quad ( V_B - V_A ) = \left [ \frac {( U_B - U_A )}{q_0} \right ] …….. (7)

Now, from equation (3) we have –

( U_B - U_A ) = - q_0 \int\limits_{B}^{A} \vec {E} . d \vec {r} ……… (3)

Therefore, potential difference is given by –

\Delta V = ( V_B - V_A ) = \left [ \frac {- q_0 \int\limits_{B}^{A} \vec {E} . d \vec {r}}{q_0} \right ]

= - \int\limits_{B}^{A} \vec {E} . d \vec {r}

__Relation between Electric Field and Potential__

Consider about a point charge ( + Q ) placed at point O as shown in figure. Points ( A ) \ \text {and} \ ( B ) are close to each other so that, electric field intensity due to charge ( + Q ) is uniform between these points.

The repulsive force experienced by a positive test charge ( + q_0 ) placed at point A will be –

\vec {F} = q_0 \vec {E} .

The direction of this force is along the direction of electric field i.e. from B \ \text {to} \ A . Work done in moving the test charge for a small distance ( dr ) from A \ \text {to} \ B will be –

dW = \vec {F} d \vec {r} = q_0 \vec {E} d \vec {r}

= q_0 E dr \cos 180 \degree = ( - ) q_0 E dr

Therefore, \quad \left ( \frac {dW}{q_0} \right ) = - E dr …….. (8)

By definition of potential difference, we will get –

dV = \left ( \frac {dW}{q_0} \right ) = - E dr

Therefore, equation (8) becomes –

dV = - E dr

So, \quad E = - \left ( \frac {dV}{dr} \right )

*Here, the negative sign indicates that electric field intensity is in the direction of decreasing electric potential gradient.*

__Equipotential Surface__

*An equipotential surface is defined as a surface of which at all points the electric potential due to a charge is same.*

Formation of equipotential surface depends upon type of *charge distribution.* Equipotential surfaces for different types of charge distributions are shown in figure.

Figure (A) shows equipotential surfaces for an uniform electric field. Here equipotential surfaces are parallel planes perpendicular to the electric field lines. Figure (B) shows equipotential surface for a positive isolated point charge. Here equipotential surfaces are concentric spherical surfaces around the isolated point charge.

Figure (C) shows equipotential surface for a pair of like (positive) point charges. Here equipotential surfaces are concentric closed surfaces around the each point charge as shown in figure. Figure (D) shows equipotential surface for an electric dipole. Here equipotential surfaces are concentric spherical surfaces originating from one end charge and terminating at other end charge of dipole.

__Properties of Equipotential Surfaces__

Equipotential surfaces have following properties –

**Work done is zero in moving a charge from one point to another point on an equipotential surface.**

We have the relation –

\Delta V = ( V_B - V_A ) = \left ( \frac {W_{AB}}{q_0} \right )

But, along an equipotential surface –

\Delta V = 0

Therefore, \quad \left ( \frac {W_{AB}}{q_0} \right ) = 0

Or, \quad W_{AB} = 0

**The electric field at any point of equipotential surface is perpendicular to the surface.**

We have the relation –

\left ( \frac {dW}{q_0} \right ) = \vec {E} . d \vec {r}

Also, work done is zero in moving a test charge on equipotential surface. Therefore, \quad \vec {E} . d \vec {r} = 0

So, \quad E dr \cos \theta = 0

Or, \quad \cos \theta = 0 \quad

Or, \quad \theta = 90 \degree

Therefore, \quad \vec {E} \perp d \vec {r}

**Equipotential surfaces are closer in strong electric field region compared to the weak electric field region.**

We have the relation –

dV = - E dr

Therefore \quad dr = - \left ( \frac {dV}{dr} \right )

So, \quad dr \propto \left ( \frac {1}{E} \right )

Therefore, if electric field is strong, the separation of equipotential surface will be less.

**Two equipotential surfaces cannot intersect to each other.**

See numerical problems based on this article.