Question

Two particles A and B, having opposite charges $2.0\times {10}^{-6}$C and $-2.0\times {10}^{-6}$C, are placed at a separation of $1.0$cm. Calculate the electric field at a point on th axis of the dipole $1.0$cm away from the centre.

Answer 1

The value of the dipole moment is :

$p=q\times d$

$p=2\times {10}^{-6}\times 0.01$

$=2\times {10}^{-8}$

We know, Electric field at an axial point of the dipole
$=\frac{2Kp}{r^3}$

$=\frac{2\times 9\times {10}^9\times 2\times {10}^{-8}}{(1\times {10}^{-2}{)}^3}$

$=36\times {10}^{7}N/C$.