Question

The dipole moment of a system of charge $+q$ distributed uniformly on an arc of radius $R$ subtending an angle $\pi /2$ at its centre where another charge $-q$ is placed is :

• A. $\frac{2\sqrt{2}qR}{\pi }$
• B. $\frac{\sqrt{2}qR}{\pi }$
• C. $\frac{qR}{\pi }$
• D. $\frac{2qR}{\pi }$

Answer 1

Answer: (A)

$\frac{2\sqrt{2}qR}{\pi }$

The definition of dipole moment is $\overrightarrow{p}=\int \overrightarrow{r}dq$
If $\lambda$ is the line charge density, charge on element $dl$ is $dq=\lambda Rd\theta$
In polar coordinate, $\overrightarrow{r}=R\cos \theta \hat{i}+R\sin \theta \hat{j}$
thus, $\overrightarrow{p}=\int (R\cos \theta \hat{i}+R\sin \theta \hat{j})\lambda Rd\theta$
$p_x=\lambda R^2{\int }_0^{\pi /2}\cos \theta d\theta =-\lambda R^2$
$p_y=\lambda R^2{\int }_0^{\pi /2}\sin \theta d\theta =\lambda R^2$
thus, $p=\sqrt{p_x^2+p_y^2}=\sqrt{2}\lambda R^2=\sqrt{2}{R}^2\frac{2q}{\pi R}$ $(q={\int }_0^{\pi /2}\lambda Rd\theta =\frac{\lambda R\pi }{2})$
$\therefore p=\frac{2\sqrt{2}qR}{\pi }$