A wheel having mass m has charges +q and -q o | vedclass

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(C) For the wheel to be in equilibrium,the net torque about the center of the wheel must be zero.The torque due to the gravitational force component $

Vedclass · Accepted Answer

$\frac{mgtan	heta}{2q}$

Vedclass · Answer

(C) For the wheel to be in equilibrium,the net torque about the center of the wheel must be zero.The torque due to the gravitational force component $mg \sin	heta$ acting at the contact point (which is at distance $r$ from the center) is $	au_g = (mg \sin	heta)r$.The electric field $E$ is vertical. The dipole moment $p = q(2r)$ is directed from $-q$ to $+q$. Let $\alpha$ be the angle between the dipole moment and the vertical electric field. From the geometry,the angle between the dipole and the horizontal is $	heta$,so the angle with the vertical is $\alpha = 90^\circ - 	heta$.The torque due to the electric field is $	au_e = pE \sin\alpha = (q \cdot 2r) E \sin(90^\circ - 	heta) = 2qrE \cos	heta$.Equating the torques: $mgr \sin	heta = 2qrE \cos	heta$.Solving for $E$: $E = \frac{mgr \sin	heta}{2qr \cos	heta} = \frac{mg 	an	heta}{2q}$.

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

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