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68 Electric Potential *P25.57 The plates create uniform electric field to the right in the picture, with magnitude VV d V d 00 0 2 −− = bg . 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 −= Ex V d x 2 0 . Its ground connection maintains it at V = 0 by allowing charge q to flow from ground onto the ball, where −+ = = 2 0 2 Vx d kq R q VxR kd e e . Then the ball feels electric force Fq E e == 4 0 2 2 to the right. For equilibrium this must be balanced by the horizontal component of string tension according to Tm g T e cos sin θθ 4 0 2 2 tan θ 4 0 2 2 kdmg x L e for small x . Then V RL e 0 2 12 4 = F H G I K J . 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 swing over to one plate or the other. P25.58 (a) Take the origin at the point where we will find the potential. One ring, of width dx , has charge Qdx h and, according to Example 25.5, creates potential
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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