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Oct20-09, 06:50 PM
P: 92
Quote Quote by turin View Post
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