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 point from the origin when `r//a gt gt 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?

Explanation

(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 from 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 `(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/((5-0)^(2)+a^(2))`
`=(-q)/(4piin_(0)sqrt(25^(2)+a^(2))+q/(4pi in_(0)sqrt(25+a^(2))`
=0
Electrostatic potentials, `V_(2)`, at point `(-7,0,0)` is given by
V_(2)=(-q)/(4pi in_(0))1/(sqrt(-7)^(2)+(-a)^(2))+q/(4pi in_(0))1/(sqrt((-7)^(2)+(a)^(2)))`
`=(-q)/(4pi in_(0))sqrt(49+a^(2))+q/(4pi in_(0))1/(sqrt(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 doen by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.