hw2sol.pdf

# hw2sol.pdf - HW II Graduate problems have an asterisk Only...

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HW II Graduate problems have an asterisk. Only those taking the course as a graduate course need to solve them. Homework from the Griffiths: Problems 2.32, 2.38, 2.52 and 2.60(*). Solution to 2.32: you have to consider both momentum and energy conservation, and possibly also the gravitational potential. The unknown v A and v B are found solving the system m A v A = m B v B , 1 2 ( m A v 2 A + m B v 2 B ) = ( q A q B 4 π 0 - Gm A m B ) 1 a . Solution to 2.38: You can find the charge surface by applying Gauss’s Law in the space between charge and shell, inside the conducting shell, and outside the conducting shell. The results are σ ( R ) = Q 4 πR 2 , σ ( a ) = - Q 4 πa 2 , σ ( b ) = Q 4 πb 2 . Knowing the potential inside a charged spherical shell, one finds immediately Φ(0) = Q 4 π 0 ( 1 R - 1 a + 1 b ) . When the outer conductor is grounded, only the outside charge will be drained, the inner charge must stay in place to insure zero electric field inside the conductor, so Φ(0) = Q 4 π 0 ( 1 R - 1 a ) .

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• Fall '15
• Giovani Bonvicini

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