Question

A thin non-conducting ring of radius $R$ has a linear charge $\lambda ={\lambda }_0\cos \theta$ where ${\lambda }_0$ is the value of $\lambda$ at $\theta =0$ . The net electric dipole moment for this charge distribution is :

• A. $\pi R^2/{\lambda }_0$
• B. ${\lambda }_0/\pi R^2$
• C. $\pi R^2{\lambda }_0$
• D. $\frac{{\lambda }_0\pi }{R^2}$

Answer 1

The correct option is C $\pi R^2{\lambda }_0$

The small length is given as,

$dl=Rd\theta$

The small charge is given as,

$dq=\lambda dl$

$dq={\lambda }_0\cos \theta Rd\theta$

The position of charge is given as,

$\overrightarrow{r}=\left(R\cos \theta ,R\sin \theta ,0\right)$

The electric dipole moment is given as,

$\overrightarrow{p}=\int \overrightarrow{r}dq$

$={\int }_0^{2\pi }{\lambda }_0\cos \theta Rd\theta \left(R\cos \theta ,R\sin \theta ,0\right)$

${\overrightarrow{p}}_x={\int }_0^{2\pi }{\lambda }_0{R}^2{\cos }^2\theta d\theta$

$=\frac{{\lambda }_0{R}^2}{2}{\int }_0^{2\pi }\left(1+\cos 2\theta \right)d\theta$

$=\frac{{\lambda }_0{R}^2}{2}{\left[\theta +\frac{1}{2}\sin 2\theta \right]}_0^{2\pi }$

$={\lambda }_0{R}^2\pi$

${\overrightarrow{p}}_y={\int }_0^{2\pi }{\lambda }_0{R}^2\cos \theta \sin \theta d\theta$

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

$=\frac{{\lambda }_0{R}^2}{2}{\left[\cos 2\theta \right]}_0^{2\pi }$

$=0$

${\overrightarrow{p}}_z=0$

Thus, net electric dipole moment is $\pi R^2{\lambda }_0$.