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EE1 HW 5 Sol

# EE1 HW 5 Sol - EE1 Spring 2008 Homework#5 Solutions(1(a...

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EE1 Spring 2008 Homework #5 Solutions (1) (a) Find the electric field E on the z-axis for z = h > 0 ? The system above can be represented by removing the conductor and replacing it with an negative line charge density of the same radius at z = -2d. To calculate the electric field for the region z = h > 0 we must add the contribution from each ring of charge. The electric field for a circle of charge on the z-axis is given by (derived from HW #2): E = ρ l az 2 0 ( a 2 + z 2 ) 3 / 2 ˆ z For each ring of charge the electric field is: E 1 = ρ l az 2 0 ( a 2 + z 2 ) 3 / 2 ˆ z 1

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E 2 = ρ l az 2 0 ( a 2 + ( z + 2 d ) 2 ) 3 / 2 ˆ z The electric field is then the sum of the two field at z=h: E tot = E 1 ( h ) + E 2 ( h ) E tot = ρ l ah 2 0 ( a 2 + h 2 ) 3 / 2 ˆ z - ρ l az 2 0 ( a 2 + ( h + 2 d ) 2 ) 3 / 2 ˆ z (b) Find the electric field ( E ) on the z-axis for z = - k where k > d Since this is inside the conductor, E = 0. 2
(2) Find the capacitance of an isolated conducting sphere of radius b that is uniformly coated with a dielectric layer of thickness d . The dielectric has an electric susceptibility of χ e We can calculate the capacitance by using the boundary condition: D 1 = D 2 , where 1 is the dielectric shell region and 2 is the free-space region outside the dielectric shell. We see that if D 1 = D 2 then r E

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