Question

Find the dipole moment of a system of charge $+q$ distributed uniformly on an arc of radius $R$ subtending an angle $\frac{\pi }{2}$ at its center where another charge $-q$ is placed.

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

Answer 1

The correct option is A $\frac{2\sqrt{2}qR}{\pi }$

The formula of dipole moment is given by $$\overrightarrow{p}=\int dq\overrightarrow{r}$$If $\lambda$ is the line charge density, so the charge on element $dl$ is $dq=\lambda Rd\theta$


In polar coordinate system:

$\overrightarrow{r}=R\cos \theta \hat{i}+R\sin \theta \hat{j}$

Thus,

$\overrightarrow{p}=\int (R\cos \theta \hat{i}+R\sin \theta \hat{j})\lambda Rd\theta$

So, the dipole moment along $x$ and $y$ axes are calculated as

$p_x=\lambda R^2{\int }_0^{\frac{\pi }{2}}\cos \theta d\theta =\lambda R^2\sin \theta {|}_0^{\frac{\pi }{2}}=\lambda R^2$

$p_y=\lambda R^2{\int }_0^{\frac{\pi }{2}}\sin \theta d\theta =\lambda R^2(-\cos \theta ){|}_0^{\frac{\pi }{2}}=\lambda R^2$

Thus, net dipole moment $p$,

$p=\sqrt{p_x^2+p_y^2}$

$\Rightarrow p=\sqrt{2}\lambda R^2$

As, total charge, $(q={\int }_0^{\frac{\pi }{2}}\lambda Rd\theta =\frac{\lambda R\pi }{2})$

$\Rightarrow p=\sqrt{2}{R}^2\times \frac{2q}{\pi R}$

$\therefore P=\frac{2\sqrt{2}qR}{\pi }$

Hence, option (a) is the correct answer.