Question

Find the magnitude of component of electric field due to a dipole at a point $\text{P}$ along perpendicular to the line joining point $\text{P}$ and centre of dipole O as shown in figure.

• A. $4.41\times {10}^{-3}\text{N/C}$
• B. $3.41\times {10}^{-3}\text{N/C}$
• C. $2.41\times {10}^{-3}\text{N/C}$
• D. $1.41\times {10}^{-3}\text{N/C}$

Answer 1

The correct option is D $1.41\times {10}^{-3}\text{N/C}$

Given:
Distance of charges from centre of dipole, $l=1\text{mm}$

Distance of point $\text{P}$ from centre of dipole, $x=4\text{m}$

Magnitude of charge, $q=10\text{nC}$

Angle made of line joining point $\text{P}$ and centre of dipole with axis of dipole, $\theta ={30}^{\circ }$

We know that magnitude of component of electric field at a point $\text{P}$ far away ($4\text{m}>>1\text{mm}$) along perpendicular to the line joining of point $\text{P}$ and centre of dipole due to a dipole is given by

$E=\frac{kp\sin \theta }{x^3}=\frac{2kql\sin \theta }{x^3}$

$\Rightarrow E=\frac{2\times 9\times {10}^9\times 10\times {10}^{-9}\times 1\times {10}^{-3}\times \sin {30}^{\circ }}{4^3}$

$\Rightarrow E=1.41\times {10}^{-3}\text{N/C}$

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