Question

What is the magnitude of electric field intensity due to a short dipole of moment of magnitude $2\times {10}^{-8}\text{C -m}$ at a point $1\text{m}$ from the centre of the dipole, when the line joining the point to the centre of the dipole makes an angle of ${60}^{\circ }$ with the dipole axis?

• A. $170\text{N/C}$
• B. $193\text{N/C}$
• C. $200\text{N/C}$
• D. $238\text{N/C}$

Answer 1

The correct option is D $238\text{N/C}$


We know that the magnitude of electric field due to a dipole at a general point $\text{P}$ in space is given by

$E=\frac{kp}{x^3}\sqrt{1+3{\cos }^2\theta }$

Here,
$p=2\times {10}^{-8}\text{C -m}$
$x=1\text{m};\theta ={60}^{\circ }$

after putting above data, we get

$\Rightarrow E=\frac{9\times {10}^9\times 2\times {10}^{-8}}{1^3}\sqrt{3(\cos 60{)}^2+1}$

$\Rightarrow E=180\sqrt{\frac{3}{4}+1}=180\sqrt{\frac{7}{4}}$

$\Rightarrow E=238\text{N/C}$

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