ch23_p47 - 47. To find an expression for the electric field...

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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 r g , concentric with the spherical shell and within it ( a < r g < b ). Gauss’ law will be used to find the magnitude of the electric field a distance r g from the shell center. The charge that is both in the shell and within the Gaussian sphere is given by the integral qd V s = z ρ 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 r dr = 4 2 π . Thus, () 22 2 2 4 4 4 2 . gg g rr r s g aa a A qr d r r d r A r d r A r a r πρ π == = = ³³ ³ The total charge inside the Gaussian surface is
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