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A point charge q of mass m is suspended vertically by a string of length l. A point dipole of dipole moment p is brought from infinity toward q such that the charge moves away from its original position. At final equilibrium the string makes some angle with the vertical, and the dipole is oriented so that the system is in static equilibrium under three coplanar forces. The work done by the external agent in bringing the dipole to this final equilibrium position equals N * (mgh), where h is the vertical rise of charge q and g is the acceleration due to gravity. Find N. (For three coplanar concurrent forces in equilibrium, F / sin(theta) is the same for all three forces, where theta is the angle between the other two.)

1. N = 1
2. N = 2
3. N = 3
4. N = 4

Correct answer: N = 2

Solution

When the dipole is brought quasi-statically to the equilibrium position, the work done by the external agent equals the change in total potential energy of the system (gravitational + electrostatic). At equilibrium, the net force on q is zero. Applying the sine rule for three concurrent forces, the electrostatic force from the dipole on q equals the component that balances the horizontal pull. Energy analysis using the work-energy theorem gives W_ext = delta_PE_grav + delta_PE_elec. At the equilibrium configuration the electrostatic PE of the dipole-charge system equals -mgh (dipole PE is negative because the dipole is aligned to attract). Hence W_ext = mgh + (-mgh) + mgh = 2mgh, giving N = 2.

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