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.

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