__What is the Electric Potential of a Dipole?__

*Electric potential of dipole a**t a point in the **electric field** produced by dipole i**s defined as the **work* *done by an external **force** in bringing a test charge from infinity to that point.*

Consider about a *dipole* AB as shown in figure. Let, –

- End charges of dipole are ( - q ) \ \text {and} \ ( + q ) .
- Distance of separation between the charges is d .
- O is the mid point of
*dipole length*i.e. ( d = AB )

Consider about a point P , where ( OP = r ) and ( \angle {POB} = \theta ) .

Assume that ( r >> d ) .

Let, AA' is a perpendicular to PO drawn from point A and BB' is another perpendicular drawn from B .

Therefore, \quad AP = A'P = ( OP + OA' ) = ( OP + AO \cos \theta ) = ( r + \frac {d}{2} \cos \theta )

Similarly, \quad BP = B'P = ( OP - OB' ) = ( r - \frac {d}{2} \cos \theta )

Potential at P due to the charge ( - q ) is –

V_1 = - \left ( \frac {1}{4 \pi \epsilon_0} \right ) \left ( \frac {q}{AP} \right )

= - \left ( \frac {1}{4 \pi \epsilon_0} \right ) \left ( \frac {q}{r + \frac {d}{2} \cos \theta} \right )

Potential at P due to the charge ( + q ) is –

V_2 = \left ( \frac {1}{4 \pi \epsilon_0} \right ) \left ( \frac {q}{BP} \right )

= \left ( \frac {1}{4 \pi \epsilon_0} \right ) \left ( \frac {q}{r - \frac {d}{2} \cos \theta} \right )

Therefore, net potential at P , due to the dipole is –

V = ( V_1 + V_2 )

= - \left ( \frac {1}{4 \pi \epsilon_0} \right ) \left ( \frac {q}{r + \frac {d}{2} \cos \theta} \right ) + \left ( \frac {1}{4 \pi \epsilon_0} \right ) \left ( \frac {q}{r - \frac {d}{2} \cos \theta} \right )

= \left ( \frac {1}{4 \pi \epsilon_0} \right ) \left [ \left ( \frac {q}{r - \frac {d}{2} \cos \theta} \right ) - \left ( \frac {q}{r + \frac {d}{2} \cos \theta} \right ) \right ]

= \left ( \frac {1}{4 \pi \epsilon_0} \right ) \left [ \frac {q d \cos \theta}{r^2 - \frac {d^2}{4} \cos^2 \theta} \right ]

= \left ( \frac {1}{4 \pi \epsilon_0} \right ) \left [ \frac {q d \cos \theta}{r^2} \right ] . Neglecting the term \left [ \frac {d^2}{4} \cos^2 \theta \right ] being very small.

So, \quad V = \left ( \frac {1}{4 \pi \epsilon_0} \right ) \left [ \frac {p \cos \theta}{r^2} \right ]

__Electric Potential of a Point Charge__

Consider about a point charge ( q ) as shown in figure. Let, P is a point at a distance ( x ) from the charge.

A unit positive charge ( + 1 ) is placed at P in the electric field produced by the charge ( q ) . The *electrostatic force* of repulsion between the charge ( q ) and charge ( + 1 ) , will be –

F = \left ( \frac {1}{4 \pi \epsilon_0} \right ) \left ( \frac {q}{x^2} \right )

= \left ( \frac {q}{4 \pi \epsilon_0 x^2} \right )

Direction of force ( \vec {F} ) is just opposite to the direction of *electric field* ( \vec {E} )

Now, let the charge ( + 1 ) is displaced by a small distance ( dx ) towards the charge ( q ) without producing any *acceleration*.

Small work done in displacing the charge will be –

dW = \vec {F} . d \vec {x}

= F . dx \cos 180 \degree = - F dx

= - \left ( \frac {q}{4 \pi \epsilon_0 x^2} \right ) dx …….. (1)

Therefore, total work done in displacing the charge from infinity to point A , will be –

\int {dW} = \int\limits_{\infty}^{r} - \left ( \frac {q}{4 \pi \epsilon_0 x^2} \right ) dx

Or, \quad W = - \left ( \frac {q}{4 \pi \epsilon_0} \right ) \int\limits_{\infty}^{ r} x^{(- 2)} dx

= - \left ( \frac {q}{4 \pi \epsilon_0} \right ) \left [ \frac {x^{- 1}}{- 1} \right ]_{\infty}^{r}

= \left ( \frac {q}{4 \pi \epsilon_0} \right ) \left [ \frac {1}{x} \right ]_{\infty}^{r}

= \left ( \frac {q}{4 \pi \epsilon_0} \right ) \left [ \frac {1}{r} - \frac {1}{\infty} \right ]

= \left ( \frac {1}{4 \pi \epsilon_0} \right ) \left ( \frac {q}{r} \right )

Because, \quad \left ( \frac {1}{\infty} \right ) = 0

But, *work done in bringing a unit positive charge from infinity to a given point = Electric potential due to the source charge at that point.*

Therefore, electric potential at a point at distance ( r ) from a point charge will be –

V_{r} = \left ( \frac {1}{4 \pi \epsilon_0} \right ) \left ( \frac {q}{r} \right )

- If the source charge is a positive i.e. ( + q ) , then electric potential ( V_{r} ) is positive.
- If the source charge is negative i.e. ( - q ) , then electric potential ( V_{r} ) is negative.

## Variation of Potential with distance

A figure is drawn to illustrate the variation of electric potential at a point with respect to its distance from the source point charge. Two graphs are most important.

- Graph of potential versus distance.
- Graph of potential versus reciprocal of distance.

### Graph of potential versus distance

- Variation of electric potential ( V ) at a point with respect to its distance ( r ) from a positive source charge is shown in figure (A).
- Variation of electric potential ( V ) at a point with respect to its distance ( r ) from a negative source charge is shown in figure (B).

The graph is having following properties.

- It is a curve.
- Potential is decreasing exponentially with increase in distance ( r )

### Graph of V versus \left ( \frac {1}{r} \right )

- Variation of electric potential V at a point with respect to its reciprocal of distance \left ( \frac {1}{r} \right ) from a positive point source charge is shown in figure (A).
- Variation of electric potential V at a point with respect to its reciprocal of distance \left ( \frac {1}{r} \right ) from a negative point source charge is shown in figure (B).

The graph has following properties.

- It is a curve.
- Potential is increasing linearly with increase in reciprocal of distance \left ( \frac {1}{r} \right ) .