Question

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

• A. $16.3\text{N/C}$
• B. $14.3\text{N/C}$
• C. $15.3\text{N/C}$
• D. $12.3\text{N/C}$

Answer 1

The correct option is A $16.3\text{N/C}$

Given:
Distance of charges from centre of dipole, $l=2\text{m};$

Distance of point $\text{P}$ from centre of dipole, $x=(2+3)\text{m}=5\text{m};$

Magnitude of charge, $q=20\text{nC}.$

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

$|{\overrightarrow{E}}_{ax}|=\frac{2k|\overrightarrow{p}|x}{(x^2-l^2{)}^2}=\frac{4kqlx}{(x^2-l^2{)}^2}$

So,
$\left|{\overrightarrow{E}}_{ax}\right|=\frac{4\times 9\times {10}^9\times 20\times {10}^{-9}\times 2\times 5}{(5^2-2^2{)}^2}$

$\Rightarrow |{\overrightarrow{E}}_{ax}|=16.3\text{N/C}$

Hence, option $\left(a\right)$ is correct.