Question

An insulating sphere of radius R has spherically symmetric but radially nonuniform distribution of charge, with a charge density ρ(r) given by
ρ(r)=ρ0[1−(5/4)(r/R)^2].

The constant ρ0 > 0 is the charge density at the center of the sphere. Note that the charge density ρ(r) is negative for 2/√5R < r 13 years ago

Answer 1

Use Gauss law to find value of field E(r).

Answer 2

You will need to use integral calculus to find charge at distance r

Answer 3

a)\[Q=\int\limits_{}^{} \rho dV\]
\[dV=4 \pi r ^{2} dr\]
Integrate from r=0 to r=R
\[Q=4 \pi \rho 0 \int\limits_{0}^{R} (1-5/4(r/R)^{2})r ^{2}dr\]
\[Q=4 \pi \rho0 R ^{3}/12\]
Charge is positive

b) As told by Vincent-Lyon, we use Gauss thm. here
A Gaussian surface is a sphere of radius r and by spherical symmetry we take the value of field to be same at all points on the sphere and RADIALLY outwards.
\[E \times 4\pi r ^{2} = \rho o(r ^{3}/3-r ^{5}/4R ^{2})/ \epsilon\]
I found the charge inside the sphere the same way as in part a)
Solve for E from here.

PLEASE TELL ME IF I AM WRONG SOMEWHERE.