Question

(a) A solid sphere, made of an insulating material, has a volume charge density of p , where r is the radius from the center of the sphere, a is constant, and a >0. What is the electric field within the sphere as a function of the radius r? Note: The volume element dv for a spherical shell of radius r and thickness dr is equal to 4tr2dr. (Use the following as necessary: a, r, and co.) magnitude E direction radially outward (b) what If? what if the charge density as a function of r within the charged solid sphere is given by ρ direction of the electric field within the sphere at radius r. (Use the following as necessary: a, r, and ) magnitudeE direction Find the new magnitude and radially outward # ) .

Answer 1

a) From Gauss's law,

$\epsilon _{0}\oint \vec{E}.\vec{ds}=\epsilon _{0}E*4\pi r^{2}=q'$ where q' is the charge enclosed by the sphere of radius r.

Now $q'=\int \rho dV=\int a/r*4\pi r^{2}dr=2\pi ar^{2}$ . Therefore $E=2\pi ar^{2}/4\pi \epsilon _{0}r^{2}=a/2\epsilon _{0}$ .

This is the electric field within the sphere as a function of radius r.

b)In this case $\rho =a/r^{2}$.

Thus $q'=\int \rho dV=\int a/r^{2}*4\pi r^{2}dr=4\pi ar$ .

Therefore $E=4\pi ar/4\pi \epsilon _{0}r^{2}=a/\epsilon _{0}r$ .