Question

.
Two equal and opposite changes of magnitude 10micro coulomb
each are fixed at a distance of 1mm from
each other to form an electric dipole calculate
the electric field intensity. (a) at a point on its
axis at a distance 1m from the center of dipole
and (ii) at a point on its perpendicular bisector
at a distance 1m from the centre of dipole​

Answer 1

Given that,

Two equal and opposite charges of magnitude is

$q=10\times10^{-6}\ C$

Distance = 1 mm

The electric dipole moment for a pair of opposite charges is defined as

The magnitude of the charge multiply by the distance between them and the direction is toward the positive charge.

We need to calculate the dipole

Using formula of dipole

$p=ql$

Where, q = charge

l = distance

Put the value into the formula

$p=10\times10^{-6}\times1\times10^{-3}$

$p=1\times10^{-8}\ Cm$

(a). We need to calculate the electric field intensity at a point on its  axis at a distance 1m from the center of dipole

Using formula of electric field intensity

$E=\dfrac{2kpr}{(r^2-l^2)^2}$

Where, p = dipole

r = distance from the center of dipole

l = distance

Put the value into the formula

$E=\dfrac{2\times9\times10^{9}\times1\times10^{-8}\times0.005}{((0.005)^2-(1)^2)^2}$

$E=90.0\times10^{-2}\ N/C$

(b). We need to calculate the electric field intensity at a point on its perpendicular bisector  at a distance 1m from the centre of dipole​

Using formula of electric field intensity

$E=\dfrac{kp}{(r+l)^\frac{3}{2}}$

Where, p = dipole

r = distance from the center of dipole

l = distance

Put the value into the formula

$E=\dfrac{9\times10^{9}\times1\times10^{-8}}{(0.005+1)^\frac{3}{2}}$

$E=89.3\ N/C$

Hence, (a). (i). The electric field intensity at a point on its  axis at a distance 1m from the center of dipole is $90.0\times10^{-2}\ N/C$

(ii). The electric field intensity at a point on its perpendicular bisector  at a distance 1m from the centre of dipole​ is 89.3 N/C