#### Question

Two charges *−q *and *+q *are located at points (0, 0, − *a*) and (0, 0, *a*), respectively.

**(a)** What is the electrostatic potential at the points?

**(b)** Obtain the dependence of potential on the distance *r *of a point from the origin when *r*/*a *>> 1.

**(c)** How much work is done in moving a small test charge from the point (5, 0, 0) to (−7, 0, 0) along the *x*-axis? Does the answer change if the path of the test charge between the same points is not along the *x*-axis?

#### Solution

**(a) **Zero at both the points

Charge − *q* is located at (0, 0, − *a*) and charge + *q* is located at (0, 0, *a*). Hence, they form a dipole. Point (0, 0, *z*) is on the axis of this dipole and point (*x*, *y*, 0) is normal to the axis of the dipole. Hence, electrostatic potential at point (*x*, *y*, 0) is zero. Electrostatic potential at point (0, 0, *z*) is given by,

`V=1/(4piin_0)(q/(z-a))+1/(4piin_0)(-q/(z+a))`

`=(q(z+a-z+a))/(4piin_0(z^2-a^2))`

`=(2qa)/(4piin_0(z^2-a^2))=p/(4piin_0(z^2-a^2))`

Where,

`in_0` = Permittivity of free space

*p* = Dipole moment of the system of two charges = 2*qa*

**(b)** Distance *r* is much greater than half of the distance between the two charges. Hence, the potential (*V*) at a distance *r* is inversely proportional to square of the distance i.e., `V prop 1/r^2 `

**(c) **Zero

The answer does not change if the path of the test is not along the *x*-axis.

A test charge is moved from point (5, 0, 0) to point (−7, 0, 0) along the *x*-axis. Electrostatic potential (*V*_{1}) at point (5, 0, 0) is given by,

`V_1=-q/(4piin_0)1/(sqrt((5-0)^2)+(-a)^2)+q/(4piin_0) 1/(sqrt((5-0)^2)+(a)^2)`

`=-q/(4piin_0sqrt(25+a^2))+q/(4piin_0sqrt(25+a^2))`=0

Electrostatic potential, *V*_{2}, at point (− 7, 0, 0) is given by,

`V=-q/(4piin_0sqrt((-7)^2+a^2))+q/(4piin_0sqrt((-7)^2+a^2))`

`V=-q/(4piin_0sqrt((49+a^2)))+q/(4piin_0sqrt((49+a^2)))`=0

Hence, no work is done in moving a small test charge from point (5, 0, 0) to point (−7, 0, 0) along the *x*-axis.

The answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.