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

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(C) 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 whe

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

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(C) 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 line joining the charges $+q$ and $-q$ is a diameter of length $2R$.Let the angle between the diameter and the inclined plane be $\alpha$. From the geometry,$\alpha = 	heta$.The torque due to gravity $(mg)$ acting at the center of mass is zero about the point of contact.The electric force on $+q$ is $qE$ (directed horizontally to the left) and on $-q$ is $qE$ (directed horizontally to the right).The torque due to the electric field about the point of contact is $	au_E = (qE \cos 	heta) R + (qE \cos 	heta) R = 2qER \cos 	heta$.However,for equilibrium on a rough plane,the torque due to gravity component $mg \sin 	heta$ acting at the center must be balanced. The torque due to the component of gravity $mg \sin 	heta$ about the contact point is $(mg \sin 	heta) R$.Equating the torques: $2qER \cos 	heta = mgR \sin 	heta$.Solving for $E$: $E = \frac{mg \sin 	heta}{2q \cos 	heta} = \frac{mg 	an 	heta}{2q}$.

$A$ 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 uniform horizontal electric field $E$. Find the value of $E$.

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