6 -6 不带电的导体球a 含有两个球形空腔,两空腔中心分别有一点电...（6-6 an uncharged conductor ball, a, contains two spherical cavities, and the center of the two cavity has a spot of electricity...）.docVIP

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6 -6 不带电的导体球a 含有两个球形空腔,两空腔中心分别有一点电...（6-6 an uncharged conductor ball, a, contains two spherical cavities, and the center of the two cavity has a spot of electricity...）.doc

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6 -6 不带电的导体球a 含有两个球形空腔,两空腔中心分别有一点电...（6-6 an uncharged conductor ball, a, contains two spherical cavities, and the center of the two cavity has a spot of electricity...）

6 -6 不带电的导体球a 含有两个球形空腔,两空腔中心分别有一点电...（6-6 an uncharged conductor ball, a, contains two spherical cavities, and the center of the two cavity has a spot of electricity...） The ball can be expressed as a potential A VA = 0 can be solved by A qA with the ball charge, then by the superposition of charged spherical surface potential, can find the potential ball A and shell B. The solution (1) from the analysis, the outer surface of a sphere A 3 * 10 8C, B in the surface of charged spherical shell - 3 x 10 - 8C, the outer surface of charged 5 * 10 8C. by the electric potential superposition of potential ball A and spherical shell B respectively. (2) the ball shell B ground and then disconnect the ball A ground ball, A charged qA, A and B potential of ball shell for Solution The ball that the outer surface of A charged 2.12 * 10 8C, from the analysis we can deduce the shell B surface charge - 2.12 x 10 - 8C, the outer surface of the charged -0.88 * 10 8C. other potential ball A and spherical shell B respectively. The conductor grounding potential of each conductor distribution changes, breaking the static balance of the original, conductor The charge will be redistributed to the establishment of new electrostatic balance. Will be charged for the Q 6-11 conductor plate A from a distance to the uncharged conductor plate near B, such as Figure (a) shows, the two conductor plate geometry identical area was S, close to two after the conductor plate distance is d (). (1) potential neglecting edge effects for two conductor plates; (2) if B is grounded, the results and how? Exercise 6 - 10 by analysis shows that the conductor plate achieves electrostatic balance, relative to the two opposite with equal charge; phase back two with equal homocharge. Then the conservation of charge can be calculated charge distribution of each conductor surface, further calculate the distribution of electric field and electric potential difference between conductors. The conductor plate B grounding po