ph1bA_08_hw7s

ph1bA_08_hw7s - Ph1b(analytic track Solution to Homework 7...

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Ph1b (analytic track): Solution to Homework 7 Chefung Chan, Winter 2008 Problem 3-3 ( 5 points ) Using Eqs.8 and 9, p.100 Purcell , we determine R so that half of Q , the induced charge on the plane, is contained within the circle of radius R : Q 2 = Z R 0 σ (2 πr ) dr = Z R 0 Qh 2 π ( r 2 + h 2 ) 3 / 2 (2 πr ) dr 1 2 = Z R 0 hrdr ( h 2 + r 2 ) 3 / 2 = ± h h 2 + r 2 ² R 0 =1 h h 2 + R 2 h h 2 + R 2 = 1 2 R = 3 h Problem 3-4 ( 5 points ) Here, we use the trick of the “method of images” described in pages 98-99 of Purcell. We replace the conducting plane with image charges in such a way that the boundary conditions (given by the potential on the conducting planes) are satisFed: 1
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y x 10cm 10cm y y -Q +Q -Q +Q ~ F = Q 2 (10 y ) 2 ˆ y + Q 2 (2 y ) 2 ( ˆ y )+ Q 2 (10 + y ) 2 ˆ y =0 1 (10 y ) 2 1 (2 y ) 2 + 1 (10 + y ) 2 =0 (10 + y ) 2 +(10 y ) 2 (10 y ) 2 (10 + y ) 2 = 1 (2 y ) 2 2(100 + y 2 ) (100 y 2 ) 2 = 1 4 y 2 y =3 . 06 cm Problem 3-8 ( 5 points ) The wire sets plates A and C to be at the same potential, which we can arbitrarily choose to be zero. That is, we are setting
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This note was uploaded on 04/28/2010 for the course PH 1b Anal taught by Professor Mckewn during the Winter '08 term at Caltech.

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ph1bA_08_hw7s - Ph1b(analytic track Solution to Homework 7...

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