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) When point P is near to charge + q,

Electrostatic potential at point P
`V_(+)=k[(q)/(r_(1))+((-q))/(r_(2))]`
`V_(+)=k[(q)/(z-a)+(q)/(z+a)]=kq[(r+a-r+a)/(z^(2)-a^(2))]`
`:. V_(+)=kqxx(2a)/(z^(2)-a^(2))`
`:. V_(+) = (kP)/(z^(2)-a^(2))[because` 2aq=P)
When point P is near to charge -q


Similarly ,
`:.` Potential at point P
`V_(-) = -(kq)/(z^(2)-a^(2))`
If any point (x, y, 0) is in xy-plane and hence it is at same distance from - q to + q on perpendicular bisector of z-axis then potential at such point is,
`V=k[(+q)/(r)+(-q)/(r)]=0`
(b) If point P is at distance of r from origin 0, then Putting z = r in equation (1) and (2),
`V=pmk(p)/(r^(2)-a^(2))`
If `(r)/(a) gt gt 1 = r gtgta ` then `a^(2)` is neglected compare with `r^(2)`
`V=pmk(P)/(r^(2))`
`:. Vprop(1)/(r^(2))`
(c) Points (5, 0, 0) and (-7, 0, 0) are on x-axis so they are on the perpendicular bisector of dipole (equatorial line) hence they are at the same distance from both the charges. As a result potential is zero.
Now if work done in moving a test charge `q_(0)`
from (5,0,0) to (-7,0,0) ,
W = `q_(0)(V_1)-V_(2))`
`= q_(0)(0)`
`:. ` W =0
No, because electric field is conservation field, the work done in moving a test charge between two point is independent of the path