Why cant electric lines of force pass through the charged sphere? Well, basically that's how a Faraday cage works, but how can it be so?

  • 2
    $\begingroup$ Have you tried reading the wikipedia page? $\endgroup$ – Kitchi Mar 1 '13 at 9:32
  • $\begingroup$ Well, i did....but the confusion grew even more.. $\endgroup$ – Ufomammut Mar 1 '13 at 15:43

An electric field line follows the direction of change of an electrostatic potential. If you choose a point where the field line start, the line will go where the electrostatic potential changes most. This is equal to the force which is exerted on a charged particle which resides at a particular point.

Example: Assume a system of two infinitely large capacitor plates. If you start your field line (you can really choose where you want to start it!) in the middle between the two capacitors, it will take the shortest possible route to one of the capacitor plates.

Back to your question/comment: If you have a conducting sphere, the potential in it is constant. Because a field line follows the direction of change, you can't have field lines there.

Mathematically, a field line $\mathbf r(t)$ for an electrostatic potential $\Phi(\mathbf r)$ is defined as

$$\frac{d\mathbf r(t)}{dt}\propto \nabla \Phi(\mathbf r(t))$$

That's not so important, but I note it for the sake of completeness.

  • $\begingroup$ This argument doesn't really answer the question if you ask me. Sure, if you assume that the potential inside of a conductor is constant, then you find that the field vanishes there because $\mathbf E = -\nabla\Phi$ by definition. But that simply begs the question: why is the potential inside of a conductor constant? In the case of electrostatics, the argument for this is relatively simply, but what about if you zap a conductor with a large spark? $\endgroup$ – joshphysics Mar 2 '13 at 5:31
  • $\begingroup$ You're right, I'm just treating the static case here, and the argument is indeed simple. My emphasis lies on the explanation of the gradient (=change). $\endgroup$ – Rafael Reiter Mar 2 '13 at 9:12

If you mean non-conducting sphere, It can not be like Faraday cage because if you put for example a point charge out of the sphere you will have non-zero electric field inside because the sphere is not able to rearrange charges on it to cancel the effect of external sources

  • $\begingroup$ But even for a conducting sphere, there wont be any flux inside it, ryte? $\endgroup$ – Ufomammut Mar 1 '13 at 15:44
  • $\begingroup$ @user21475, charged spheres and conducting spheres are very different. Which one do you mean? $\endgroup$ – Emilio Pisanty Mar 1 '13 at 23:31
  • $\begingroup$ What if i say, charged conducting sphere? $\endgroup$ – Ufomammut Mar 2 '13 at 4:31
  • $\begingroup$ @user21475: A conducting sphere is only charged at its boundary and neutral everywhere else per definition, see e.g. my literature reference physics.stackexchange.com/a/53458/19945 $\endgroup$ – Rafael Reiter Mar 2 '13 at 9:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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