0
$\begingroup$

The following problem and its solution is taken from I. E. Irodov's book basic laws of electromagnetism :

 i.e. irodov's book basic laws of electromagnetism Relevant figure

I do not understand how the fact that field is axisymmetric leads to the conclusion that flux through sphere is same as the cylinder circumscribing it.

As charge inside cylinder is more than that in sphere , by gauss's law shouldn't it ( the flux) be more?

I found another way to solve the problem - solving the integral $$ \oint E\cdot dS=\oint E*n*dS $$ where n is the normal vector for sphere that is $$n= \frac{(x\widehat{i} +y\widehat{j}+z\widehat{k})}{\sqrt{x^2 + y^2 + z^2}}$$ then the answer is same as in the book (this method was new to me , I found it on some russian site).

So it is correct. Hence I think I'm misinterpreting the argument in the book. Please help me understand it.

So as we know flux through cylinder = that through sphere , using gauss's law , that would imply that there is no net charge in the volume difference between sphere and cylinder. How is that possible?

Can someone tell me how the charge distribution in this case looks like?

$\endgroup$
1
$\begingroup$

The electric field in the problem has no $z$ component, so it quite simple to calculate the flux through a cylinder with axis parallel to the $ z $ axis; then you choose a cylinder that contains the sphere you are interested in. Let $\Sigma$ be the surface of the cylinder, $ V $ its volume, $\Sigma '$ and $ V' $ the surface and volume of the sphere; by the divergence theorem:

$$ \int_{\Sigma} \mathbf{E} d\mathbf{S} = \int_{V} \text{div}(\mathbf{E}) dV $$

By the additivity of the integral:

$$ \int_{V} \text{div}(\mathbf{E}) dV = \int_{V'} \text{div}(\mathbf{E}) dV' + \int_{V''} \text{div}(\mathbf{E}) dV'' $$ where $V''$ is the region you obtain subtracting $V'$ from $ V$;

but on $ V'' $, $ \text{div}(\mathbf{E})=0$ since this region contains no charge; thus you get:

$$ \int_{\Sigma} \mathbf{E} d\mathbf{S} = \int_{V} \text{div}(\mathbf{E}) dV = \int_{V'} \text{div}(\mathbf{E}) dV =\int_{\Sigma '} \mathbf{E} d\mathbf{S} $$

and thus the flux through the cylinder it is equal to the flux through the sphere.

$\endgroup$
12
  • $\begingroup$ Why does V'' contain no charge? That is precisely what was bugging me. $\endgroup$ – A Googler Feb 13 '15 at 18:21
  • $\begingroup$ @AGoogler The first time I read I understood the whole charge was contained in the sphere; actually it is not clear from the text. $\endgroup$ – NNec Feb 13 '15 at 18:32
  • $\begingroup$ The radius R is arbitrary so I don't think its only in the sphere. Further , if it were only in the sphere I don't think its formula would be like that in the question. Also the solution using integrals that i gave doesn't assume that yet gets the same answer. $\endgroup$ – A Googler Feb 13 '15 at 18:42
  • $\begingroup$ @AGoogler In any case, the vector normal to the sphere surface is $ x \mathbf{i} + y \mathbf{j} + z \mathbf{k} $, while that normal to the lateral surface of the cylinder is $ x \mathbf{i} + y \mathbf{j} $; since the electric field has no $ z$ component, its internal product with those 2 vectors is the same and hence the flux. $\endgroup$ – NNec Feb 13 '15 at 19:00
  • $\begingroup$ So using gauss's law , that would imply that there is no net charge in the volume difference between sphere and cylinder. How is that possible? $\endgroup$ – A Googler Feb 13 '15 at 19:06

Not the answer you're looking for? Browse other questions tagged or ask your own question.