# solutionhw9 - Solutions to Homework Set 9 1 Conducting...

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Solutions to Homework Set 9 1) Conducting strip : A ( k ) = Z + -∞ V ( x ) e - ikx dx = V 0 Z a - a e - ikx dx = 2 V 0 sin ka k From this we have V ( x, 0 , y ) = 2 V 0 Z -∞ dk 2 π sin( ka ) k e ikx e -| k || y | . Taking the y gradients to get E y , and then getting σ = 0 ( E y | y =+ ε - E y | y = - ε ) gives us σ ( x ) = 4 V 0 0 Z -∞ dk 2 π sgn( k ) sin( ka ) e ikx e - ε | k | . The integral is elementary, and gives σ ( x ) = 4 V 0 0 4 π 1 x + a - - 1 x - a - + 1 x + a + - 1 x - a + We can take safely take the ε regulator to zero. We end up with σ ( x ) = 2 0 V 0 π 1 a + x + 1 a - x = 4 0 V 0 π a a 2 - x 2 . σ x a -a A sketch of the charge distribution on the strips. 2) Qual Problem : a) We have, in general b l = 2 l + 1 2 R l +1 Z 1 - 1 d (cos θ ) V ( θ ) P l (cos θ ) . 1

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Plugging in the explicit expression for the relevent P l gives b 1 = 3 4 R 2 ( V 1 - V 2 ) b 2 = 0 (1) b 3 = 7 16 R 4 ( V 2 - V 1 ) b) If V ( r, θ, φ ) = 1 4 π 0 Q r + d · ˆ r r 2 + · · · , then Q is the total charge, and d is the dipole moment. Thus our divided sphere has | d | = 3 π 0 R 2 ( V 1 - V 2 ) . c) For the sphere immersed in the external field we have V ( r, θ, φ ) = -| E | r cos θ + V sphere ( r, θ, φ ) , and so V ( r, θ, φ ) = -| E | r cos θ + 1 4 π 0 Q r + | d | cos θ r 2 + · · · . The surface of the sphere at r = R must be an equipotential, and so the coefficients of the cos θ terms must add to zero: -| E | R + | d | 4 π 0 R 2 = 0 .
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