The electric field is radial, so the flux through the Gaussian surface is Φ=42πrEg, where Eis the magnitude of the field. Gauss’ law yields 420222ππεErqA ragg=+−di. We solve for E: EqrAAar−LNMMOQPP14220222πππFor 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. With a= 2.00 ×10−2 m and q= 45.0 ×10−15 C, we have 1121.79 10C/m .A−=×49. To find an expression for the electric field inside the shell in terms of Aand the distance from the center of the shell, select Aso 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
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