Question

Find the net electric field at an equatorial point $\text{P}$ of a dipole as shown in figure.

• A. $1.3\times {10}^9\text{N/C}$
• B. $2.3\times {10}^9\text{N/C}$
• C. $3.3\times {10}^9\text{N/C}$
• D. $4.3\times {10}^9\text{N/C}$

Answer 1

The correct option is D $4.3\times {10}^9\text{N/C}$

Given:
Distance of charges from centre of dipole, $l=3\text{mm};$
Distance of point $\text{P}$ from centre of dipole, $x=4\text{mm};$
Magnitude of charge, $q=10\mu \text{C};$

We know that the net electric field due to a dipole at an equatorial point is given by

$|{\overrightarrow{E}}_{eq}|=\frac{k|\overrightarrow{p}|}{(x^2+l^2{)}^{3/2}}=\frac{k\left(q2l\right)}{(x^2+l^2{)}^{3/2}}$

$\Rightarrow |{\overrightarrow{E}}_{eq}|=\frac{9\times {10}^9\times 10\times {10}^{-6}\times 2\times 3\times {10}^{-3}}{{10}^{-9}\times (4^2+3^2{)}^{3/2}}$

$\Rightarrow |{\overrightarrow{E}}_{eq}|=4.3\times {10}^9\text{N/C}$

Hence, option (d) is the correct answer.