Question

An electric dipple has dipple moment $4\times {10}^{-11}$ and the two charges have magnitudes 4nC each. Calculate the electric field due to the dipole
i) At a point on the axis at a distance 0.01m from one of the charges and
ii) At a point on the perpendicular bisector at a distance 0.02m from the middle point.

Answer 1

Dipole moment $\left(P\right)=4\times {10}^{-11}$

Charge $\left(q\right)=4nC$

Length of the dipole $\left(d\right)=\frac{p}{q}=\frac{4\times {10}^{-11}}{4\times {10}^{-9}}={10}^{-2}m$

$\Rightarrow$ Radius $=\frac{1}{\frac{100}{2}}=5\times {10}^{-3}$

(i) Electric field due to dipole on the axial line.

$E=\frac{1}{4x{\epsilon }_0}\frac{2P}{{\pi }^3}$

Here $\pi =30+0.01=50.0/m$

$\Rightarrow E=9\times {10}^9\times \frac{2\times 4\times {10}^{-11}}{(5\times {10}^{-3}{)}^3}=\frac{72\times {10}^{-2}}{125075.015\times {10}^{-9}}$

$=\frac{72\times {10}^{-2}}{1.2\times {10}^{-7}}=54.4\times {10}^{5}N{C}^{-1}=57\times {10}^{5}N{C}^{-1}$

(ii) Electric field due to dipole on equitorial line

$E=\frac{1}{4\pi {\epsilon }_0}\frac{P}{{\pi }^3}$

Here $\pi =0.02$

$\Rightarrow E=9\times {10}^9\times \frac{4\times {10}^{-11}}{(0.02{)}^3}=\frac{36\times {10}^{-2}}{8\times {10}^{-6}}=4.5\times {10}^{4}N{C}^{-1}$