qual_19_p2_sol.pdf - QUALIFYING EXAMINATION Part 2 Solutions Problem 1 Electromagnetism I(a Since our system has spherical symmetry the electric field

qual_19_p2_sol.pdf - QUALIFYING EXAMINATION Part 2...

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QUALIFYING EXAMINATION, Part 2 Solutions Problem 1: Electromagnetism I (a) Since our system has spherical symmetry the electric field and potential have, respec- tively, the form E = E ( r ) ˆ r and ϕ = ϕ ( r ). We use the integral form of Gauss’s law to determine the electric field. For r < R , we have E ( r )(4 πr 2 ) = Q enc ε 0 = - q ε 0 4 3 πr 3 4 3 πR 3 = - q ε 0 r R 3 , or E ( r ) = - q 4 πε 0 R 3 r . For r > R , we find E ( r ) = - q 4 πε 0 1 r 2 . Since E = -∇ ϕ , we can find the electrostatic potential by integrating the electric field. Up to an integration constant, the potential is given by ϕ ( r ) = - R E r ( r ) dr . For r > R , using the boundary condition φ ( r ) 0 for r → ∞ , we find ϕ ( r ) = - q 4 πε 0 1 r . For r < R , we find ϕ ( r ) = + 1 2 q 4 πε 0 R 3 r 2 + C , where C is an integration constant. We determine C using the continuity condition for the electrostatic potential at r = R , and find C = - 3 2 q (4 πε 0 R ) - 1 . Therefore, the potential for r < R is ϕ ( r ) = - 1 2 q 4 πε 0 R " 3 - r R 2 # .
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