A nonconducting ring of mass m and radius R, with charge per unit length `lambda` is shown in fig. It is then placed on a rough nonconducting horizontal plane. At time t=0, a uniform electric field `vecE =E_(0)hati` is switched on and the ring starts rolling without sliding. Determine the friction force (magnitude and direction ) acting on the ring.

Answer

`lamnda RE _0 hat I`.

Explanation

As ` E` will trey rotate the ring is clockwese direction friction will uppose the rotation and provide traslatonal mtion. So frictional force will act froward .
`d tau = (dq) xx (E_0) 2 R sin theta`
`d tau = ( lambda R d theta) 2 R e_0 sin theta`

`tau = 2 R^2 lambda E_0 int_0^(pi//2) sin theta d theta = 2 R ^2 lambdaE_0` ..... (1)
For translatiional motion
` f_r Ma rArr a = f//M` ...(2)
For rotatinal motiaon
` rau - f-r =(MR^2) (a//r)` ...(3)
From (1) and (4)
` f-r = (tau) /(2Q) = (2R^2 almbdaE_0)/(2R) = R lambda E rArr E vec f_r = R lambdaE_0 hat i`.