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I have positively charged hollow spherical shell at origin, with radius R. Then negatively charged dot charge at (2R, 0, 0), and I'm supposed to find electric field everywhere.

What seems unclear is the field "behind" the sphere, where from you can not draw straight line to negative point charge. As in the question it is not stated if the spherical shell is conductor material or not. Isn't it with conductor materials that, the electric field of the dot charge can not penetrate through sphere, and thus not contributing to net field everywhere, where it would have to go through the sphere?

And also if the material of spherical shell is conductor, doesn't the point charge change how the postitive charge is distributed at it's surface, which makes it more complicated to calculate the electric field?

It seems to me that it would be too complicated getting the electric field, if the shell would be conductor, and that it should just be assumed to be insulator, but I'd like to be sure.

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I don't have enough rep to simply comment, but it's problems like these that Gauss' Law excels at solving.

Following procedure, you draw a Gaussian surface that encloses your sphere and has some symmetries that allow the integral to simplify (in your case a sphere of radius greater than R).

Physically speaking, inside this sphere, the E-field should be zero as there is no charge enclosed for r < R by Gauss' Law.

Outside the sphere r > R we should see the Electric field of a simple point charge of Q [ if we say the sphere has charge distribution Q/Area --> charge enclosed is (Q/Area)*Area ].

Conductor or insulator, the charge within your Gaussian surface should be the same as long as your surface encloses the whole charged sphere, so the distribution shouldn't matter.

I hope this helped. Comments to improve my answer are very welcome.

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  • $\begingroup$ I can post comments on my own answer, so please anyone feel free to discuss here. $\endgroup$
    – Jordan Simba
    Commented Mar 4, 2016 at 19:57
  • $\begingroup$ Clear for field of sphere with uniform charge distribution. But what if the charge distribution is not uniform? Because isn't that what is caused by the dot charge, if the material of sphere is conductor? $\endgroup$
    – PyHop
    Commented Mar 4, 2016 at 20:23
  • $\begingroup$ And also can that then be somehow applied to situation with both spherical shell and dot charge? As I've thought electric field cannot go through conductor, which (I guess) would mean the net field is not just sum of both everywhere outside the sphere? Or am I completely wrong now? $\endgroup$
    – PyHop
    Commented Mar 4, 2016 at 20:33
  • $\begingroup$ If you wish to find out the E-field of your sphere (on its own in free-space), we can only measure the effect of the field, hence we introduce a test charge, and in order for our solution to be that of JUST the sphere, we ignore the effects of the test charge on the sphere. So you were right when you asked your question. $\endgroup$
    – Jordan Simba
    Commented Mar 5, 2016 at 22:36

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