Question

A nonconducting ring of mass $m$ and radius $R$ is charged as shown. The charged density i.e. charge per unit length is $\lambda$ . It is then placed on a rough nonconducting horizontal surface plane. At time $t=0$ , a uniform electric field $\vec{E}=E_0\hat{i}$ is switched on and the ring starts rolling without sliding. The friction force (magnitude and direction) acting on the ring, when it starts moving is

• A. $\lambda R{E}_0\hat{i}$
• B. $3\lambda R{E}_0\hat{i}$
• C. $\sqrt{2}\lambda R{E}_0\hat{i}$
• D. $\frac{2}{\lambda R{E}_0}\hat{i}$

Answer 1

Answer: (A)

As $\vec{E}$ will try to rotate the ring is clockwise direction friction will oppose the rotation and provide translational motion. so frictional force will act forward.

$d\tau =(dq)\times \left(E_0\right)2R\sin \theta$
$d\tau =(\lambda Rd\theta )2R{E}_0\sin \theta$ $\tau =2R^2\lambda E_0\underset{0}{\overset{\pi /2}{\int }}\sin \theta d\theta =2R^2\lambda E_0$
For translational motion,
$f=ma\Rightarrow a=\frac{f}{m}$
For rotational motion, $\tau -fR=\left(m{R}^2\right)\left(\frac{a}{R}\right)$
From (2) and (3)
$\tau =2f_{r}R$ .....(4)
From (1) and (4)
$f_r=\frac{\tau }{2R}=\frac{2R^2\lambda E_0}{2R}=R\lambda E_0\Rightarrow \vec{f}r=R\lambda E_0\hat{i}$