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hw01_solutions

# hw01_solutions - Solutions to Problem Set 1 1 The potential...

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Solutions to Problem Set 1 1. The potential on the axis z is Φ( z ) = 1 4 π 0 Z S σdS r = 1 4 π 0 Z a 0 2 πρdρ p z 2 + ρ 2 = σ 2 0 h p z 2 + a 2 - | z | i By symmetry, E points along the axis away from the disc (for σ > 0), E = E z e z , thus E z = - Φ( z ) ∂z = ± σ 2 0 1 - | z | z 2 + a 2 where the plus sign corresponds to z > 0 while the minus sign for z < 0. 2. The conductor and the metal plate have the same potential when they are contacted. After the first contact, q/C 1 = ( Q - q ) /C 2 . After the final contact, there is no charge transfer, so q f /C 1 = Q/C 2 . The ultimate charge is q f = qQ Q - q . 3. We start with the Poisson equation, 2 Φ = - ρ 0 . The potential has no angular dependence, therefore, ρ = - 0 1 r d 2 dr 2 ( r Φ) = - q 4 πr d 2 dr 2 h e - αr 1 + αr 2 i = - 3 8 π e - αr This is valid for r > 0. For r 0, the potential Φ diverges and also ρ diverges. Note that in this limit lim r 0 Φ( r ) = q 4 π 0 r Because 2 1 r = - 4 πδ ( r ) we have to add this piece to the solution. Finally, we get ρ ( r ) = ( r ) - 3 8 π e - αr 4. a) The potential satisfies the Poisson equation Φ 00 ( x ) = - ρ/ 0 . The boundary conditions are Φ(0) = 0 and Φ 0 ( d ) = σ 0 / 0 . The solution is Φ( x ) = - ρ 2 0 x 2 + ρd + σ 0 0 x. b) The threshold shift is Φ(
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