Question

part 1 of 3 Consider a solid insulating sphere of radius b with nonuniform charge density p = ar, where a is a constant. Find the charge contained within the radius r<b as in the figure. The volume element dV for a spherical shell of radius r and thickness dr is equal to 47 r2 dr. part 2 of 3 If a = 5 x 10-6 C/m' and b = 1 m, find E at r = 0.6 m. The permittivity of a vacuum is 8.8542 x 10-12 C2/N.m”. Answer in units of N/C.

Answer 1

$The\ charge \ will \ be given\ by \ Q =\int {\rho}dV=\int{ar}\ (4{\pi}r^2)dr= \int(4{\pi}a)r^3dr=({\pi}ar^4)$

2. To determine the Electric field, first draw the Gaussian surface passing through the point where we want to find the electric field. Then gauss law states$\int(E. dA) = {Q_{inside}}/{\epsilon}$.

E = $\dfrac{Q}{4{\pi}{\epsilon}r^2}$ $=\dfrac{{\pi}ar^4}{4{\pi}{\epsilon}r^2}= \dfrac{ar^2}{4{\epsilon}}$$=\dfrac{(5\times10^{-6})\times(0.6)^2}{4\times(8.854\times 10^{-12})}= 5.1\times10^4 N/C$

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