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Charge inside Conductor

Above there is a 2D pic of this problem.

$S$ is a conducting sphere with no charge. I am considering the electrostatics case.

It is a hollow sphere: inside its cavity lies a point charge $q$, $q > 0$.

What is the electrostatic force $\vec{F}$ on the point charge $q$?

My attempt:

If $\partial S $ is border of the cavity, I know there is a total charge of $-q$ on it (because $S$ is a conductor).

Now, $\vec{F} = q \vec{E}$, where $\vec{E}$ is the electric field on the charge $q$ caused by the charge $-q$ on $\partial S$.

The problem is now about $\vec{E}$. $\vec{E} = 0$ inside the cavity if no charge is inside the cavity. What if there is $q$ inside it?

Because of symmetry, I thought that $\vec{E} = 0$ as well: there is no "main direction" the electric field should have. Neither do the force on the charge. And I also thought that the electric field on every point inside the cavity should be zero as well. Whereas it would be non-zero if charge if moved and the symmetry is lost.

However, I couldn't find a rigorous way to prove it. If I consider a Gauss surface inside the cavity, the flux is $> 0$ because $\frac{q}{\epsilon_0} > 0$, so why should the electric field be zero?

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closed as off-topic by John Rennie, ZeroTheHero, Emilio Pisanty, Kyle Kanos, user191954 Nov 14 '18 at 8:15

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  • $\begingroup$ "the flux is > 0". why do you conclude this? $\endgroup$ – garyp Nov 7 '18 at 22:51
  • $\begingroup$ @garyp $\Phi_{\Sigma} (\vec{E}) = \frac{q}{\epsilon_0} > 0$ because $q > 0$, where $\Sigma$ is the gaussian surface around $q$ and inside the cavity. $\endgroup$ – moonknight Nov 7 '18 at 22:55
  • $\begingroup$ Yes, I'm sorry, I was typing faster than I was thinking. However, I think you should be focusing on the force on the charge, not the total field. $\endgroup$ – garyp Nov 7 '18 at 22:59
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If I consider a Gauss surface inside the cavity, the flux is $>0$ because $\frac{q}{\epsilon_0}>0$, so why should the electric field be zero?

There is a difference between the field at the location of the charge $q$ and the field at another point in the cavity. Indeed, you are correct that by symmetry $E=0$ at the charge $q$ by charges on the outside of the cavity. However, if you are looking at a Gaussian sphere centered on $q$, then you are looking at the field caused by $q$.

Since this is a homework problem I will leave it to you to apply Gauss's law inside the cavity. But you can reason that the field in the cavity must be radial centered on $q$.

You can also use superposition. You already said that $E=0$ inside of the cavity without a charge in it. So then what is the field inside the cavity with the charge if we know superposition is valid for electric fields?

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  • $\begingroup$ I think there's a fine point here that needs clarification. If you are looking at a Gaussian sphere centered on $q$, the net flux through that sphere is still the flux due to all charges, not merely the flux caused by the charge $q$. $\endgroup$ – garyp Nov 7 '18 at 23:21
  • $\begingroup$ @garyp Actually if you think about it, the field due to charges on the outside is $0$ anyway, so you could argue that the outer charges don't contribute to the flux at all right? I guess it depends on when you add up the contributions from the outer charges: before or during the integral. $\endgroup$ – Aaron Stevens Nov 7 '18 at 23:30
  • $\begingroup$ I suppose you could argue that way. My point of view has always been that Gauss' Law applies to all charges and all fluxes, and the fact that charges outside don't contribute is a consequence. But I see your point. $\endgroup$ – garyp Nov 7 '18 at 23:54
  • $\begingroup$ @garyp I agree, you do have to be careful. In general you are right that everything needs to be considered. It's just in this specific case the field from all of the outer charges cancels out. Thanks for pointing this out though. $\endgroup$ – Aaron Stevens Nov 8 '18 at 0:07

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