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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» Solution :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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