ch23-p049

# ch23-p049 - 49. To find an expression for the electric...

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The electric field is radial, so the flux through the Gaussian surface is Φ= 4 2 π rE g , where E is the magnitude of the field. Gauss’ law yields 42 0 22 2 ππ ε Er q A r a gg =+ di . We solve for E : E q r A Aa r L N M M O Q P P 1 4 2 2 0 2 2 2 π π π For the field to be uniform, the first and last terms in the brackets must cancel. They do if q – 2 π Aa 2 = 0 or A = q /2 π a 2 . With a = 2.00 × 10 2 m and q = 45.0 × 10 15 C, we have 11 2 1.79 10 C/m . A 49. 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 r g , concentric with the spherical shell and within it ( a < r g < b ). Gauss’ law will be used to find the magnitude of
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