Answer:
Part a)
[tex]\rho = 3\epsilon_0 k r^2[/tex]
Part b)
[tex]Q = 4\pi \epsilon_0kR^5[/tex]
Explanation:
Part a)
As we know that electric field intensity due to some given charge distribution is given as
[tex]E = kr^3 \hat r[/tex]
now electric flux through a spherical surface of radius r is given as
[tex]\phi = E. A[/tex]
[tex]\phi = kr^3(4\pi r^2)[/tex]
now by Guass law we know that
[tex]E.A = \frac{q}{\epsilon_0}[/tex]
[tex]q = 4\pi \epsilon_0kr^5[/tex]
now volume charge density is given as
[tex]\rho = \frac{q}{\frac{4}{3}\pi r^3}[/tex]
[tex]\rho = 3\epsilon_0 k r^2[/tex]
Part b)
Total charge inside the radius R is given as
[tex]Q = 4\pi \epsilon_0kR^5[/tex]