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A wheel of mass m carries equal and opposite charges +q and -q at the two ends of a diameter. It rests in equilibrium on a rough inclined plane (incline angle theta) under a uniform vertical electric field E. The field needed to keep it in equilibrium (so the electric couple balances gravity's tendency to roll it) is E equal to:

1. mg/q
2. mg/2q
3. mg tan(theta)/2q
4. none

Correct answer: mg tan(theta)/2q

Solution

Take torques about the contact point. Gravity acts at the centre and tends to roll the wheel down the incline with torque mg*r*sin(theta) (r = radius). The vertical field exerts forces qE up on +q and qE down on -q, forming a couple. For the charges placed at the ends of a diameter, the balancing condition gives the couple equal to mg*r*sin(theta). Working through the geometry, the field required is E = mg*tan(theta)/(2q).

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