Differential form of Gauss's Law

  • Thread starter Anti-Meson
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  • #1
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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 :

[tex]\int_{S} \underline{E}.\underline{dS} = \int_{V} \frac{\rho}{\epsilon_{0}} dV [/tex]

Yields a result [tex]\rho = \frac{3E\epsilon_{0}}{r} [/tex] for 0<r<=a

Homework Statement



However when using the differential form:

[tex]\nabla . \underline{E} = \frac{\rho}{\epsilon_{0}}[/tex]

[tex]\frac{1}{r} \frac{\partial (r E_{r})}{\partial r} = \frac{\rho}{\epsilon_{0}} [/tex]

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

[tex] \rho = {\epsilon_{0}} E (1 + ln a ) [/tex]


Any helpful advice would be appreciated.
 
Last edited:

Answers and Replies

  • #2
turin
Homework Helper
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I think you accidently used the cylindrical version of the divergence.

Why are you integrating?
 
  • #3
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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,

[tex] \frac{\partial (r E_{r})}{\partial r} = E_{r} [/tex]

or if i use a radial divergence from the beginning:

[tex] \nabla . E = \frac{\partial (E)}{\partial r} [/tex]

but by using that I get:

[tex] E = \frac{\rho r}{\epsilon_0} [/tex]

i.e missing a factor of 3
 
Last edited:

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