**What is Electrostatic Potential?**

The amount of work done to move a unit charge from a reference point to a specific point inside the field without producing an acceleration.

**Electric Potential Formula**

= | electric potential | |

= | Coulomb constant | |

= | charge | |

= | distance of separation |

**SI Unit of Electrostatic Potential**

The SI unit of electrostatic potential is volt.

SI unit of electrostatic potential | volt |

Other units | statvolt |

Symbol of electrostatic potential | V or φ |

Dimensional formula | ML^{2}T^{3}I^{-1} |

We know that central forces are conservative in nature i.e., work done on any particle moving under the influence of conservative forces does not depend on path taken by the particle but depends on initial and final positions of the particle.

Electrostatic force given by Coulomb’s law is also a central force like gravitational force and is conservative in nature.

For conservative forces, work done on particle undergoing displacement can be expressed in terms of the potential energy function.

**Electrostatic Potential**

Work done per unit test charge is characteristic of the electric field associated with the charge configuration. This leads to the idea of electrostatic potential V due to a given charge configuration.

Work done by external force in bringing a unit positive charge from point R to P

where V_{P} and V_{R} are the electrostatic potentials at P and R, respectively.

Work done by an external force in bringing a unit positive charge from infinity to a point = electrostatic potential (V) at that point.

**In other words,** the electrostatic potential (V) at any point in a region with electrostatic field is the work done in bringing a unit positive charge (without acceleration) from infinity to that point.

**SI unit of Potential difference is Volt. 1V=1Nm C ^{-1}.**

**Potential Due to a Point Charge**

Consider a point charge Q at the origin. For Q > 0, the work done against the repulsive force on the test charge is positive. Choose a convenient path along the radial direction from infinity to the point P.

At some intermediate point P′ on the path, the electrostatic force on a unit positive charge is

where rˆ′ is the unit vector along OP′. Work done against this force from r′ to r′ + ∆r′ is

The negative sign appears because for ∆r′ < 0, ∆W is positive. Total work done (W) by the external force is obtained by integrating from r′ = ∞ to r′ = r,

This, by definition, is the potential at P due to the charge Q

Figure below shows how the electrostatic potential ( ∝ 1/*r*) and the electrostatic field ( ∝ 1/*r ^{2}*) varies with r.

**Potential Due to an Electric Dipole**

Take the origin at the center of the dipole. Since the potential is related to the work done by the field, the electrostatic potential also follows the **superposition principle**. Thus, the potential due to the dipole is the sum of potentials due to the charges q and –q.

where r_{1} and r_{2} are the distance of the point from q and –q, respectively.

Now, by geometry,

Take r much greater than a (r ≫ a) and retain terms only up to the first order in a/r

Similarly,

Using the Binomial theorem and retaining terms up to the first order in a/r; obtain,

Using Equations and p = 2qa, we get

Now,

where is the unit vector along the position vector OP. The electric potential of a dipole is then given by

The equation is approximately true only for distances large compared to the size of the dipole so that higher order terms in a/r are negligible. For a point dipole p at the origin,

From Eq. the potential on the dipole axis (θ= 0, π) is given by

**(Positive sign for θ= 0, negative sign for θ= π.) The potential in the equatorial plane (θ= π/2) is zero.**

- The potential due to a dipole
**depends**not just on r but also on the angle between the position vector r and the dipole moment vector p. - The electric dipole potential falls off, at large distance, as 1/r
_{2}, not as 1/r, characteristic of the potential due to a single charge

### Potential due to a system of charges

Consider a system of charges q_{1}, q_{2},…, q_{n} with position vectors r_{1}, r_{2},…, r_{n} relative to some origin. The potential V_{1} at P due to the charge q_{1} is

where r_{1P} is the distance between q_{1 } and _{P}. Similarly, the potential V_{2} at P due to q_{2} and due to q are given by

where r_{2P} and r_{3P} are the distances of P from charges q_{2 }and q_{3}, respectively; and so on for the potential due to other charges.

By the superposition principle, the potential V at P due to the total charge configuration is the algebraic sum of the potentials due to the individual charges

The electric field outside the shell is as if the **entire charge is concentrated at the centre**. Thus, the potential outside the shell is given by

where q is the total charge on the shell and R its radius. The electric field inside the shell is zero. This implies that potential is constant inside the shell (as **no work is done in moving a charge inside the shell**), and, therefore, equals its value at the surface, which is