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 (0, 0, z) and(x, y, 0) ? (b) Obtain the dependence of potential on the distance r of a pointfrom the origin when r/a >> 1. (c) How much work is done in moving a small test charge from thepoint (5,0,0) to (–7,0,0) along the x-axis? Does the answerchange if the path of the test charge between the same points is not along the x-axis?

Answer 1

Given: The locations of two charges are ( 0,0,−a ) and ( 0,0,a ).

a)

The electrostatic potential at point ( 0,0,z )is given as,

V= 1 4π ε 0 ( q z−a )+ 1 4π ε 0 ( −q z+a )

Where, the permittivity of free space is ε 0 , the charge is −q at the location ( 0,0,−a ) and the charge is +q at the location ( 0,0,a ) a point location

Further simplify the above expression, we get

V= q( z+a−z+a ) 4π ε 0 ( z 2 − a 2 ) = 2qa 4π ε 0 ( z 2 − a 2 ) = p 4π ε 0 ( z 2 − a 2 )

Where, the dipole moment of the system of two charges is p.

Thus, the electrostatic potential at the points is p 4π ε 0 ( z 2 − a 2 ) .

b)

Since, the distance r is much greater as compared to the half of the distance between the two charges therefore,

The potential at distance r is given as,

V∝ 1 r 2

Thus, the potential at a distance r is inversely proportional to square of the distance.

(c)

Given: The locations of the given charges are ( 5,0,0 ) and ( −7,0,0 ).

The electrostatic potential at point ( x,0,0 )is given as,

V 1 = −q 4π ε 0 × 1 ( x−0 ) 2 + ( −a ) 2 + q 4π ε 0 × 1 ( x−0 ) 2 + ( a ) 2

By substituting the given values in the above expression, we get

V 1 = −q 4π ε 0 × 1 ( 5−0 ) 2 + ( −a ) 2 + q 4π ε 0 × 1 ( 5−0 ) 2 + ( a ) 2 = −q 4π ε 0 × 1 25+ a 2 + q 4π ε 0 × 1 ( 25 )+ a 2 = −q 4π ε 0 25+ a 2 + q 4π ε 0 25+ a 2 =0

The electric potential at point ( x,0,0 ) is given as,

V 1 = −q 4π ε 0 × 1 ( x−0 ) 2 + ( −a ) 2 + q 4π ε 0 × 1 ( x−0 ) 2 + ( a ) 2

By substituting the given values in the above expression, we get

V 2 = −q 4π ε 0 × 1 ( −7−0 ) 2 + ( −a ) 2 + q 4π ε 0 × 1 ( −7−0 ) 2 + ( a ) 2 = −q 4π ε 0 × 1 49+ a 2 + q 4π ε 0 × 1 49+ a 2 = −q 4π ε 0 49+ a 2 + q 4π ε 0 49+ a 2 =0

Thus, no work is done in moving a small test charge from point ( 5,0,0 )to ( −7,0,0 ) . The work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points so the answer will not change.