- #1
gavman
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Homework Statement
A hollow spherical shell carries charge density [itex]\rho=\frac{k}{r^2}[/itex] in the region [itex]a<=r<=b[/itex], where a is the inner radius and b is the outer radius. Find the electric field in the region [itex]a<r<b[/itex].
I'm not allowed to use integral form of Gauss's law, must use differential form.
Homework Equations
Relating charge density to electric field divergence [itex]\vec{∇}\cdot\vec{E}=\frac{\rho}{\epsilon_{0}}[/itex]
The Attempt at a Solution
Using the integral method, I believe the electric field is [itex]\vec{E}(\vec{r})=\frac{k}{\epsilon_{0}}\frac{r-a}{r^2}\hat{r}[/itex]
I then went ahead and determined that spherical co-ordinates are the way to go, and that the E-field has only an [itex]\hat{r}[/itex] component and put together
[itex]\vec{∇}\cdot\vec{E}=\frac{1}{r^2}\frac{∂(r^2\vec{E}(\vec{r}))}{∂r}=\frac{k}{r^2\epsilon_{0}}[/itex]
By inspection I decided that I should have [itex]\vec{E}(\vec{r})=\frac{k}{\epsilon_{0}r}\hat{r}[/itex]. While this does satisfy [itex]\vec{∇}\cdot\vec{E}=\frac{\rho}{\epsilon_{0}}[/itex], it is not in agreement with the integral form.
So my question is, am I using these relationships incorrectly? I'm very confused, since when I integrate ∇E from a to r, I get the same answer as the Gauss integral form, but not sure why this could be, since I understand that the integral of ∇E is equal to the charge enclosed by a Gaussian surface. Since I'm just interested in the E-field at a point in the shell, why would I need to integrate anyway?