Symmetric Charge Distribution

For the spherically symmetric charge distribution located in free space, having $\rho=\left\{\begin{matrix}3r^2&\text{ for r<a}\\\rho=0&\text{ for r>a}\end{matrix}\right.$

a) Use the potential definition and integrate the charge distribution to find the potential at the center of the sphere.

b) Find the electric field from Gauss law then find the potential by integrating the field over a path from $0$ to $\infty$.


At the center of the distribution $r =0$ so that ($dV=r^2*dr*\sinθ dθ*dφ$  and $k=1/4πε$):

$ϕ(0)=∫ k\frac{ρ(r’)}{r}’dV’=k\left(∫_0^a \frac{ρ(r)}{r}r^2dr\right)\left(∫_0^π \sinθ dθ\right)\left(∫_0^{2π} dφ\right)= 4πk∫_0^a 3r^3dr=3πka^4$


From local Gauss law one has for $0<r<a$ (inside the sphere)


And in spherical coordinates

$∇f= \frac{1}{r^2} *\frac{d}{dr} (r^2*f)$  

since $E=E(r)$  given the spherical distribution of charge so that

$\frac{d}{dr} (r^2E)=\frac{ρr^2}{ε}  (=\frac{3r^4}{ε})$  or

$r^2E(r)=∫ \frac{3r^4}{ε}dr=\frac{3r^5}{5ε}+C$


Outside the sphere $R>a$ the Gauss law is

$E(R)*S(R)=Q_{inside}/ε$    with     $Q_{inside}=∫ ρ*dV$
$Q_{inside}=∫_0^a 3r^2*r^2*dr∫_0^π \sinθ dθ*∫_0^{2π} dφ=4π*(3a^5)/5$

so that

$E(R)=\frac{4π*(3a^5)/5}{4πR^2 ε}=\frac{3a^5}{5R^2 ε}$

At $r=R=a$ the filed need to be continuous.

$(3a^3)/5ε+C=(3a^3)/5ε$    so that $C=0$

The potential at the center of the sphere is (we exchange limits $∞→0$   with $0→∞$ and change sign in front of integral:

$ϕ(0)=∫_0^a E_{inside}*dr+∫_a^∞ E_{outside}*dr=3/5ε*∫_0^a r^3*dr+(3a^5)/5ε ∫_a^∞ 1/R^2 *dR$


which is the same result as above.