Two large parallel vertical conducting plates are separated d so that their potentials are +V and –V. A small conducting ball is hung midway of radius R and mass m (where R<<d) is hung midway. The thread of length L supporting the thread is conducting wire connected to ground, so that the potential of the ball is fixed at V = 0. Show that the equilibrium of the ball when V is sufficiently small. Show that the equilibrium is unstable if V exceeds value
ked^2mg/(4RL)

This problem is rather subtle. First, notice that there is no charge on the ball initially, so the force on it is zero. Hence it is in equilibrium. The question is whether this equilibrium is stable or unstable.

There are two forces at work here. Since the ball is hanging from a wire of length L, when it is displaced infinitesimally to the right by an amount dx, the wire will tilt slightly to an angle theta = dx/L. The vertical component T_y of the tension in the wire will balance against gravity, so T_y = mg, and the horizontal component T_x pulls the ball back towards the origin, producing a stable equilibrium. T_x = T*sin(theta) and T_y = T*cos(theta) = mg, so T_x = mg*tan(theta) = mg*tan(dx/L) = mg/L*dx, since tan(theta) = theta for small angles. Therefore, the force T_x acts exactly like a spring with spring constant k = mg/L. (This is a standard formula for a pendulum. You could just look it up in any first year physics textbook, but I am giving the derivation here so the discussion will be complete.)

The second force is trickier to think about. Although there is no charge on the ball initially, if you displace it to the right by a small amount dx, it will no longer be sitting at the location where V=0. The potential at its new location will be (2V/d)*dx, since V varies linearly between the two large plates. But at the same time, the ball is still connected to ground through a wire, so its potential MUST be 0. The only way that this can happen is if the ball sucks a small amount of negative charge up through the wire, giving itself enough negative potential to compensate (by the superposition principle) for the positive potential it is sitting in.

How much charge is needed to give a conducting ball of radius R a potential V? By Gauss's law, the electric field produced by such a ball will be the same as from a point charge, and so the potential at the ball's surface will be the same as the potential a distance R from a point charge. This is k_eQ/R. Hence k_eQ/R = -(2V/d)dx. Since the ball now has a charge, it will feel an electric force attracting it towards the positive plate (i.e. the force is in the +x direction). This force will be F = EQ = (2V/d)(2V/d*dx*R/k_e) = (2V/d)^2*R/k_e*dx, which again is proportional to dx, so it acts like a spring. Since this "spring" is pushing away from the origin, it acts like it has a negative spring constant k = -(2V/d)^2*R/k_e.

When the spring constant pushing towards the origin is stronger than the spring constant pushing away from it, the equilibrium will be stable, and in the opposite case, unstable. The border between the two cases is then obviously when the two spring constants are equal in strength, i.e. mg/L = (2V/d)^2*R/k_e. Solving this for V gives V^2 = k_ed^2mg/(4RL). This is exatly the answer you want, but this is V^2. Did you make a mistake writing V instead of V^2?

A word of warning about this problem: The use of Gauss's Law here is unjustified, because the field from the large plates will cause the charge on the small sphere to be distributed NON-UNIFORMLY, so the symmetries used by Gauss's law to get the E field from a sphere will not be true here. Even if the sphere is very small, so that it can be treated as a point charge when seen from far away, the surface potential is calculated by integrating the E field from infinity to the surface, and much of the contribution to the total value of the integral comes from the very strong field near the surface, where it is highly non-uniform.. This means that the approximation that this ball acts like a point charge is probably not valid even if R is extremely small. In other words, this problem is probably actually much harder that it seems. The book's answer is almost certainly wrong, unless the effects of the uneven surface charge distribution on the sphere somehow cancel out, and I cannot see why they would.

Hope this helps!

--Stuart Anderson

Thank you very much for your detailed solution.

You are welcome! I posted this almost exactly a year ago, and when I saw the date, at first I didn't notice the year and thought it was from a few days ago, but I didn't remember posting anything like this. They I noticed the 2007. How time flies!

--Stuart Anderson