A wheel of mass m has charges +q and -q at di | vedclass

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(B) For the wheel to be in equilibrium, the net torque about the point of contact with the inclined plane must be zero.Let $r$ be the radius of the wh

Vedclass · Accepted Answer

$\frac{mg 	an 	heta}{2q}$

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(B) For the wheel to be in equilibrium, the net torque about the point of contact with the inclined plane must be zero.Let $r$ be the radius of the wheel. The gravitational force $mg$ acts at the center of the wheel. The torque due to gravity about the point of contact is $	au_g = (mg \sin 	heta)r$.The electric field $E$ is vertical. The dipole moment $p = q(2r)$ is directed from $-q$ to $+q$. From the geometry, the angle between the dipole moment and the vertical electric field is $	heta$.The torque due to the electric field is $	au_e = pE \sin 	heta = (2qr)E \sin 	heta$.Equating the torques: $(mg \sin 	heta)r = (2qr)E \sin 	heta$.Solving for $E$: $E = \frac{mg \sin 	heta}{2q \sin 	heta} = \frac{mg}{2q}$.Thus, the correct option is $B$.

$A$ wheel of mass $m$ has charges $+q$ and $-q$ at diametrically opposite points. It remains in equilibrium on a rough inclined plane of inclination $\theta$ in the presence of a vertical electric field $E$. Find the value of $E$.

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