InterviewSolution
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InterviewSolution
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As shown in (Fig. 3.90), two large parallel vertical conducting plates separated by distance D are charge so that their potential are `+V_0` and `- V_0`. A small conducting ball of mass m and radius r `("where" R lt lt d)` is hung midway between the plates. The thread of length L supporting the ball is a conducting wire connected to ground, so the potential of the ball is fixed at `V = 0`. The ball hangs straight down in stable equilibrium when `V_0` is sufficiently small. Show that the equilibrium of the ball is unstable if `V_0` exceeds the critical value `[k_e d^2 mg//(4 R L)]^(1//2)`. |
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Answer» The plates create uniform electric field to the right in the picture, with magnitude `[V_0 - (-V_0)]// d = 2V_0 //d`. Assume the ball swings a small distance x to the right. It moves to a place where the voltage created by the plates is lower by `- E x = -(2 V_0)/d x` Its ground connection maintains it at `V = 0` by allowing charge q to flow from ground onto the ball, where `- (2V_0 x)/d + (k_e q)/R = 0` `q = (2 V_0 x R)/(k_e d)` Then the ball feels electric force `F = q E = (4 V_0^2 x R)/(k_e d^2)` to the right. for equilibrium, this must be balanced by the horizontal component of string tension according to `T cos theta = mg` `T sin theta = (4 V_0^2 x R)/(k_e d^2) tan theta = (4 V_0^2 x R)/(k_e d^2 mg) = (x)/(L)` for small x Then `V_0 = ((k_e d^2 m g)/(4 R L))^(1//2)` If `V_0` is less than this value, the only equilibrium position of the ball is hanging straight down. If `V_0` exceeds this value, the ball will saving over to one plate or the other. |
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