P24_045 - 45 To find an expression for the electric field...

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Unformatted text preview: 45. To find an expression for the electric field inside the shell in terms of A and the distance from the center of the shell, select A so the field does not depend on the distance. We use a Gaussian surface in the form of a sphere with radius rg , concentric with the spherical shell and within it (a < rg < b). Gauss’ law will be used to find the magnitude of the electric field a distance rg from the shell center. The charge that is both in the shell and within the Gaussian sphere is given by the integral qs = ρ dV over the portion of the shell within the Gaussian surface. Since the charge distribution has spherical symmetry, we may take dV to be the volume of a spherical shell with radius r and infinitesimal thickness dr : dV = 4πr2 dr . Thus, r r r g g g A2 2 qs = 4π ρr2 dr = 4π r dr = 2πA(rg − a2 ) . r dr = 4πA r a a a 2 The total charge inside the Gaussian surface is q + qs = q + 2πA(rg − a2 ). The electric field is radial, so 2 the flux through the Gaussian surface is Φ = 4πrg E , where E is the magnitude of the field. Gauss’ law yields 2 2 4πε0 Erg = q + 2πA(rg − a2 ) . We solve for E : E= 2πAa2 1 q + 2πA − 2 2 4πε0 rg rg . For the field to be uniform, the first and last terms in the brackets must cancel. They do if q − 2πAa2 = 0 or A = q/2πa2 . ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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