Answer :

The charges –q and + q at (0,0,-a) and (0,0,a) respectively forms a dipole.

(a) The point (0,0,z) lies on the axis of dipole, while the point (x,y,0) lies on the equatorial plane of the dipole.

The electric potential of dipole on any point on its axis is given by the formula–

where, = Absolute Permittivity of free space

p = Dipole moment of the system of two charges

r = Distance of point from the centre of the dipole

here, p = 2qa and r = z.

The electric potential of dipole at point (0,0,z) =

The electric potential of dipole on any equatorial point is zero

The electric potential of dipole at point (x,y,0) = 0.

(b) Given r/a>>1, which implies r>>a

the distance of point where potential is to be obtained 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 1/r2

(c) A test charge is moved from point (5, 0, 0) to point (-7, 0, 0) along the x-axis.

Electrostatic potential (V1) at point (5, 0, 0) is given by,

V1 =


= 0

Electrostatic potential(V2) at point (-7, 0, 0) is given by,

V2 =


= 0

The potential difference between points (5,0,0) and (-7,0,0,) is V1-V2 = 0.


The work done in moving a small test charge from the point (5,0,0) to (–7,0,0) along the x-axis = 0.

The answer does not change if the path of the test charge between the same points is not along the x-axis 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.

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