(a) Derive the expression for the electric potential at any point P, at distance r from the centre of an electric dipole, making angle , with its axis. (b) Two point cha

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

Answers (1)

a)
consider an electric dipole of length 2a. The point p is at a distance r from the centre of the dipole making an angle d with the centre of the dipole.

From Geometry
$r_{2}= r+a\cos d$
$r_{1}= r-a\cos d$
Potential at

$p,V= \frac{kq}{r_{1}}- \frac{kq}{r_{2}}$
                          $= kq\left [ \frac{1}{r_{1}}- \frac{1}{r_{2}} \right ]$
                         $= kq\left [ \frac{1}{r-a\cos d}- \frac{1}{r+a\cos d} \right ]$
                        $= \frac{kq2a\cos d}{r^{2}-a^{2}\cos ^{2}d}$
      $V= \frac{kp\cos d}{r^{2}-a^{2}\cos ^{2}d}$
For    $r> > > a$
Potential $V= \frac{kp\cos d}{r^{2}}$
where    $k= \frac{1}{4\pi \varepsilon _{0}}$
$\therefore V= \frac{p\cos d}{4\pi \varepsilon _{0}r^{2}}$
b)

Let the electric field is zero at p at distance x from $1\mu c$ charge, then
$\frac{k\times 4\mu c}{\left ( 2-x \right )^{2}}= \frac{k\times 1\mu c}{x^{2}}$
$\frac{2}{2-x}= \frac{1}{x}$
$2x= 2-x$
$x= \frac{2}{3}m$
$\therefore 2-x= 2-\frac{2}{3}= \frac{4}{3}m$