Point charge inside hollow conducting sphere [closed]

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?

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

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• "the flux is > 0". why do you conclude this? – garyp Nov 7 '18 at 22:51
• @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. – moonknight Nov 7 '18 at 22:55
• 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. – garyp Nov 7 '18 at 22:59

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?
• 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 $q$. – garyp Nov 7 '18 at 23:21
• @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. – Aaron Stevens Nov 7 '18 at 23:30