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 whe

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$\frac{mg}{2q}$

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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 line connecting the charges $+q$ and $-q$ is a diameter of length $2R$.The gravitational force $mg$ acts downwards at the center of the wheel. The component of gravitational force along the incline is $mg \sin 	heta$,which creates a torque about the point of contact equal to $(mg \sin 	heta) R$.The electric field $E$ is vertical. The force on $+q$ is $qE$ upwards and the force on $-q$ is $qE$ downwards.The perpendicular distance from the point of contact to the line of action of the electric forces is $R \sin 	heta$.The torque due to the electric field about the point of contact is $(qE) \cdot (R \sin 	heta) + (qE) \cdot (R \sin 	heta) = 2qER \sin 	heta$.Equating the torques: $2qER \sin 	heta = mgR \sin 	heta$.Solving for $E$,we get $E = \frac{mg}{2q}$.

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

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