- #1
roam
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Homework Statement
Find the electric field inside and outside a sphere of radius R using the differential form of Gauss's law. Then find the electrostatic potential using Poisson's equation.
Charge density of the sphere varies as ##\rho (r) = \alpha r^2 \ (r<R)## and ##\rho(r)=0 \ elsewhere##.
Homework Equations
##\nabla . E = \frac{\rho}{\epsilon_0}##
##\nabla^2 \phi = - \frac{\rho}{\epsilon_0}## (Poisson)
The Attempt at a Solution
I'm confused since using the differential form seems redundant, because you must first find the electric field using the integral form (unless I'm mistaken). Here's my attempt:
For inside the sphere (##r<R##) the charge enclosed is:
##Q=\rho V = \alpha r^2 . \frac{4}{3} \pi r^3 = \frac{4\alpha r^5 \pi}{3}##
Divergence of the field in spherical coordinates is:
##\nabla . \vec{E} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 E_r) + \frac{1}{r sin \theta} \frac{\partial}{\partial \theta} (sin \theta E_\theta) + \frac{1}{r sin \phi} \frac{\partial}{\partial \phi} ##
Since no change in ##\theta## and ##\phi## we can just take the first term:
##\nabla . \vec{E} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 E_r)##
Now finding ##E_r##
##|E_r| \oint da = |E_r| 4 \pi r^2 = \frac{4 \alpha r^5 \pi}{3 \epsilon_0} ##
##E_r = \frac{\alpha r^3}{3 \epsilon_0}##
So substituting back:
##\nabla . \vec{E} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 \frac{\alpha r^3}{3 \epsilon_0})##
##=\frac{\alpha}{3 \epsilon_0} \frac{\partial}{\partial r} r^3 = \frac{\alpha r^2}{\epsilon_0}= \frac{\rho(r)}{\epsilon_0}##
##=\frac{\alpha}{3 \epsilon_0} \frac{\partial}{\partial r} r^3 = \frac{\alpha r^2}{\epsilon_0}= \frac{\rho(r)}{\epsilon_0}##
Now integrating we get:
##\int(\nabla. E) dr=\int \frac{\alpha r^2}{\epsilon_0} dr = \frac{\alpha r^3}{3 \epsilon_0} + C##
So it's the same answer as using the integral form of Gauss's law. So what is the point of using the differential form?
Now using Poisson's equation:
##\nabla^2 \phi = - \frac{\rho}{\epsilon_0} = - \frac{\alpha r^2}{\epsilon_0}##
I think I must take a double integral:
##\int \int - \frac{\alpha r^2}{\epsilon_0} dr dr##
##\frac{-\alpha}{\epsilon_0} \int (\frac{r^3}{3} + C) dr##
##\phi = \frac{-\alpha}{\epsilon_0} (\frac{r^4}{12}+Cr+B)##
##\frac{-\alpha}{\epsilon_0} \int (\frac{r^3}{3} + C) dr##
##\phi = \frac{-\alpha}{\epsilon_0} (\frac{r^4}{12}+Cr+B)##
Is this right? And what I should I do about the constants?
And when we are outside the sphere (r>R), with ##\rho=0## we will have ##Q=(0)V=0##. Therefore both the electric field and ##\phi## will be zero?
Any help would be greatly appreciated.