Question

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

• A. $3.5\times {10}^{-3}\text{N/C}$
• B. $2.0\times {10}^{-3}\text{N/C}$
• C. $3.0\times {10}^{-3}\text{N/C}$
• D. $2.5\times {10}^{-3}\text{N/C}$

Answer 1

The correct option is D $2.5\times {10}^{-3}\text{N/C}$

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

Distance of point $\text{P}$ from centre of dipole, $x=5\text{m}$

Magnitude of charge, $q=10\text{nC}=10\times {10}^{-9}\text{C}$

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

We know that the magnitude of an electric field at a point $\text{P}$ far away ($5\text{m}>>1\text{mm}$) along the line joining of point $\text{P}$ and centre of dipole due to a dipole is given by

$E=\frac{2kp\cos \theta }{x^3}=\frac{4kql\cos \theta }{x^3}$

$\Rightarrow E=\frac{4\times 9\times {10}^9\times 10\times {10}^{-9}\times 1\times {10}^{-3}\times \cos {30}^{\circ }}{5^3}$

$\Rightarrow E=2.5\times {10}^{-3}\text{N/C}$

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