(a) Derive the expression for the electric potential at any point P, at distance r from the centre of an electric dipole, making angle `alpha`, with its axis.
(b) Two point charges `4 mu C and + 1 mu C` are separated by a distance of 2m in air. Find the point on the line joining charges at which the net electric field of the system is zero.

Explanation

(a) `V_(1)` = Potentil at P due to `(-q) = (1)/(4 pi epsi_(0)) xx (-q)/(r_(1))`
`V_(2)=` Potential at P due to `q = (1)/(4 pi epsi_(0)) xx (q)/(r_(2))`
Potential at `P = V = V_(1) + V_(2)`
`V = (-1)/(4pi epsi_(0)) xx (q)/(r_(1)) + (1)/(4pi epsi_(0)) xx (q)/(r_(2))`
`rArr V = (q)/(4pi epsi_(0)) [(1)/(r_(2)) - (1)/(r_(1))]`
`r_(1) = AP = CP = r + 1 cos theta and r_(2) BP = DP = r - l cos theta`
`V = (q)/(4pi epsi_(0)) [(1)/((r - l cos theta)) - (1)/((r + l cos theta))]`

`rArr V = (q)/(4 pi epsi_(0)) [(r + l cos theta r + l cos theta)/(r^(2) - l^(2) cos^(2) theta)]`
`V = (q)/(4pi epsi_(0)) [(2l cos theta)/(r^(2) - l^(2) cos^(2) theta)]`
`V = (2l q cos theta)/(4pi epsi_(0) (r^(2) - l^(2) cos^(2) theta)) = (P cos theta)/(4pi epsi_(0) (r^(2) - l^(2) cos^(2) theta))`
(b) `E = (1)/(4pi epsi_(0)) (q)/(r^(2)) rArr E_(1) = (1)/(4pi epsi_(0)) xx (4 xx 10^(-6))/(x^(2)) and E_(2) = (1)/(4pi epsi_(0)) xx (1 xx 10^(-6))/((2 -x)^(2))`
ATQ `E_(1) = E_(2) or (1)/(4 pi epsi_(0)) xx (4 xx 10^(-6))/(x^(2)) = (1)/(4pi epsi_(0)) xx (4 xx 10^(-6))/((2 - x)^(2))`
`(2)/(x) = (1)/(2 -x) rArr 4 - 2x = x " " x = .^(4)//_(3)m and 2-x = .^(2)//_(3) m`