Here, -q is at (0,0, -a) and +q is at (0,0,a)
(i) Potential at `(0,0,z)` would be
`V = (1)/(4pi in_(0)) ((-q)/(z+a)) + (1)/(4pi in_(0)) (q)/((z-a)) = (q(z+a-z+a))/(4pi in_(0)(z^(2) - a^(2))) = (q2a)/(4pi in_(0)(z^(2) - a^(2))) = (p)/(4pi in_(0) (z^(2) - a^(2)))`
Potential at (x,y,0), i.e., at a point `_|_` to z-axis where charges are located, is zero.
(ii) We have proved that `V = (p cos theta)/(4pi in_(0) (r^(2) - a^(2) cos^(2) theta))`
If `(r)/(a) gt gt 1`, then `a lt lt r :. V = (p cos theta)/(4pi in_(0) r^(2)) :. V prop (1)/(r^(2))`
i.e., potential is inversely proportional to square of the distance.
(iii) Potential at `(5,0,0) is V_(1) = (-q)/(4 pi in_(0)) (1)/(sqrt((5-0)^(2) + (-a)^(2))) + (q)/(4pi in_(0)) (1)/(sqrt((5-0)^(2) + a^(2)))`
`= (-q)/(4pi in_(0) sqrt(25+a^(2))) + (q)/(4pi in_(0) sqrt(25+a^(2))) = zero`
Potential at `(-7,0,0) is V_(2) = (-q)/(4pi in_(0)) (1)/(sqrt((-7 - 0)^(2) + a^(2))) + (q)/(4pi in_(0)) (1)/(sqrt((-7 - 0)^(2) + a^(2))) = zero`.
As work done = charge `(V_(2) - V_(1)) :.` W = zero.
As work doen by electrostaic field is independent of the path connecting the two points, therefore, work doen will continue to be zero along every path.