Question

A charge particle of charge q and mass m is clamped rigidly at the circumference of a ring with mass m and radius R. Initially ring is in vertical plane resting on a sufficiently rough horizontal surface with charge q at the same horizontal level as that of the centre of the ring. There exist uniform horizontal electric fields as shown. At$t=0$ the system is let free. Given that $(qE=mg)$. Using the above information answer the following. (Assume there is no sliding at point of contact at any moment of time during motion).
Work done by the electric field when the ring has rotated through 90 (use $\pi =\frac{22}{7}$)

• A. $\frac{2}{7}mgR$
• B. $\frac{4}{7}mgR$
• C. $-mgR$
• D. $mg\sqrt{2}R$

Answer 1

Answer: (B)

$\frac{4}{7}mgR$

The ring rotates by ${90}^o$. Thus the charge which is on the circumference will move a distance $\frac{\pi }{2}R$ around the circumference. Also the whole sphere has moved a distance of R in forward direction.
Thus we get the horizontal displacement of particle till ring as
$=dx=\frac{\pi }{2}R-R=R(\frac{22}{7\times 2}-1)$
$dx=\frac{4}{7}R$
Work done by the electric field, $W=F.dx=qE.\frac{4}{7}R=\frac{4}{7}mgR,(mg=qE)$