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Imagine there is a cylinder with a charge density of +Q per unit volume and of infinite length. Now place a spherical cavity inside it with a diameter equal to the cross-section diameter of the cylinder. Is there an electric field inside the sphere? If so, is it possible to calculate the E-field with Gauss's Law?

a spherical cavity inside cylinder

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Yes, you can use Gauss's law, but I will leave you to work out the details. You use the principle of superposition.

Use Gauss's law (cylindrical symmetry) to work out the E-field inside the uniform cylinder, without the spherical hole in it.

Use Gauss's law (spherical symmetry) to work out what the E- field would be due to a sphere with a negative charge density $-Q$, in the position you have shown the spherical cavity.

Your situation is equivalent to the sum of these two fields.

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  • $\begingroup$ Thank you very much for your answer. But I'm confused little bit. As far as I know, with Gauss' Law, we can only find the electric field on a determined surface (gaussian surface). How is it possible to calculate the sum of all electric fields inside the whole sphere by Gauss' Law? There are infinitely many gaussian surfaces in the sphere. $\endgroup$ – ogulcan Oct 7 '16 at 18:12
  • $\begingroup$ Yes, and each has its own radius. The E-field is a fn of radius. $\endgroup$ – Rob Jeffries Oct 7 '16 at 21:39
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I do not see any symmetry which would make Gauss law useful here. There is a symmetry plane separating the upper and the lower cylinder and there is rotational symmetry around the cylinder axis. Considering these symmetries you'll probably have to use Coulomb's law for the electric fields and integrate over all charge elements in the cylinder.

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  • $\begingroup$ Thanks a lot. But could you please explain why there is an electric field inside the sphere although the upper and the lower parts of the plane are symmetries of each other? Shouldn't the electrical field be zero? $\endgroup$ – ogulcan Oct 4 '16 at 21:05
  • $\begingroup$ Except for the center of the sphere, where the electric field has to be zero, there should be an electric field in the symmetry plane pointing radially outwards. $\endgroup$ – freecharly Oct 4 '16 at 22:46
  • $\begingroup$ Rob Jeffries has pointed out in his answer a way how to make use of Gauss law in this case, which is probably the easiest way to get the field distribution. $\endgroup$ – freecharly Oct 4 '16 at 22:48

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