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I know that, in the center of a uniformly charged hollow sphere, the electric field strength E will be zero.

Exact question: If one penetrates a uniformly charged hollow sphere, then the electric field strength E

(a) increases

(b) decreases

(c) remains the same as at the surface

(d) is zero at all points

Though I have tried a lot, I cannot understand it. I think if a uniformly charged hollow sphere is penetrated, there will be some electric field line in the center. Am I right or wrong? What is the answer of the above problem? Please, Explain the logic behind it.

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closed as off-topic by StephenG, freecharly, Qmechanic May 22 '18 at 4:43

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  • $\begingroup$ Define what you mean by penetrated ? $\endgroup$ – StephenG May 21 '18 at 22:50
  • $\begingroup$ There will be a hole or a pore in the surface of the hollow sphere. $\endgroup$ – MARSHAL SHAWKAT May 21 '18 at 22:55
  • $\begingroup$ I think the proper word there is "perforates," rather than "penetrates." $\endgroup$ – probably_someone May 21 '18 at 23:48
  • $\begingroup$ In the book, the word is "penetrate" not "perforate". $\endgroup$ – MARSHAL SHAWKAT May 22 '18 at 0:03
  • $\begingroup$ @MARSHALSHAWKAT In that case, you might be misinterpreting the problem. If one penetrates a uniformly charged hollow sphere, then it only means that one moves through it, from the outside to the inside. Penetration does not necessarily mean that one creates a hole; for example, a rock can penetrate the surface of a lake, but it does not leave a hole in the surface. $\endgroup$ – probably_someone May 22 '18 at 0:58
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Since the charge is uniform, they will be symmetrically arranged and the field vectors will cancel inside the sphere (at all points). In this case, the symmetry is the only thing that matters.

If you think of a more general case, where a bunch of charges are free to move on a conductor closed surface, they will move around until all forces are balanced. In this equilibrium situation, the electric field inside the conductor is also zero.

In the sphere case, if the charges were free to move along the surface, they would be uniformly arranged when in equilibrium.

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  • $\begingroup$ but what's about the pore/hole in the sphere? $\endgroup$ – MARSHAL SHAWKAT May 21 '18 at 22:57
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If you're talking $\mod{E}$ then it will have to increase (the lowest value it can take is 0. There's no other option for it. I suspect that due to the symmetry there's probably still be a significant portion of it at 0 though. And for problems like this the calculations become rapidly complex and require numerical analysis rather. If the whole is small enough you can approximate that it's not there and then it would be essentially 0.

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  • $\begingroup$ I can not understand the line "... still be a significant portion of it at 0". $\endgroup$ – MARSHAL SHAWKAT May 22 '18 at 0:06
  • $\begingroup$ Just my rough guess that even if you put a big hole in it the field would probably still be 0 at a lot of places $\endgroup$ – Jake Rose May 22 '18 at 0:07
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According to Newton's shell theorem, which also hold for an electrostatic field caused by charges according to Coulomb's law, the electric field inside a homogeneously charged sphere must be zero. Thus, when you penetrate the surface of the uniformly charged sphere the electrical filed inside the sphere will go from a maximum value at the surface to zero inside the surface.

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