Differential form of Gauss's Law

[Anti-Meson]

Homework Statement

In a model of an atomic nucleus, the electric field is given by:

E = α r for r < a

where α is a constant and a is the radius of the nucleus.


Use the differential form of Gauss's Law to calculate the charge density ρ inside the nucleus.


2. The attempt at a solution

Using the simple version of Gauss's law :

$$\int_{S} \underline{E}.\underline{dS} = \int_{V} \frac{\rho}{\epsilon_{0}} dV$$

Yields a result $$\rho = \frac{3E\epsilon_{0}}{r}$$ for 0<r<=a

Homework Statement

However when using the differential form:

$$\nabla . \underline{E} = \frac{\rho}{\epsilon_{0}}$$

$$\frac{1}{r} \frac{\partial (r E_{r})}{\partial r} = \frac{\rho}{\epsilon_{0}}$$

and when integrating with respect to r from 0 to a,

$$\rho = {\epsilon_{0}} E (1 + ln a )$$


Any helpful advice would be appreciated.

 

[turin]

I think you accidently used the cylindrical version of the divergence.

Why are you integrating?

 

[Anti-Meson]

I think you accidently used the cylindrical version of the divergence.

Why are you integrating?

I have used the cylindrical version, though this problem only depends on r so maybe use spherical? I integrated the partial differential with respect to r.

if I assume that Er is not a function of r but a constant then,

$$\frac{\partial (r E_{r})}{\partial r} = E_{r}$$

or if i use a radial divergence from the beginning:

$$\nabla . E = \frac{\partial (E)}{\partial r}$$

but by using that I get:

$$E = \frac{\rho r}{\epsilon_0}$$

i.e missing a factor of 3