What happens when one of two cocentric spherical shells is grounded?

[Davidllerenav]

Homework Statement

A conducting spherical shell has charge Q and radius R1. A larger
concentric conducting spherical shell has charge −Q and radius
R2. If the outer shell is grounded, explain why nothing happens to
the charge on it. If instead the inner shell is grounded, find its final
charge.

Relevant Equations

$V=-\int_\infty ^a \vec E\cdot d\vec l$

Hi! I need help with this problem.
When the outer shell is grouded, its potential goes to zero, $V_2=0$ and so does it charge, right? $-Q=0$. So the field would be the one produced by the inner shell $E=\frac{Q}{4\pi\epsilon_0 R_1^2}$.

When the inner shell is grounded, I think that something similar would happen, $V_1=0$ and $Q=0$, but I think it is wrong, because the problem asks me to find the final charge of the inner shell. I'm pretty confused about what happens when a conductor is grounded.

 

[haruspex]

When the outer shell is grouded, its potential goes to zero,

Yes, but what was its potential initially?

-Q=0

Why?

and Q=0

Again, why?
The potential at a point is the sum of potentials due to all charges present. If one shell has no charge then its potential is that due to the charge on the other shell.

 

[Davidllerenav]

Yes, but what was its potential initially?

I think that it would be zero, right? Becasue hte field is zero, since the charge enclosed would be Q-Q.

The potential at a point is the sum of potentials due to all charges present. If one shell has no charge then its potential is that due to the charge on the other shell.

When the inner shell is grounded, I need to calculate the potential from infinity to 0, right? and since when R2<r<R1 the field is zero because the Q charge is gone, then the potential would be $V=\frac{-Q}{a\pi\epsilon_0 R_2^2}$, right? Same thing would happen when the outer shell is grounded I think.

 

[Orodruin]

Grounding something means its potential is set to zero. This does not imply that the charge becomes zero - it becomes whatever it needs to be for the potential to be zero.

 

[Davidllerenav]

Grounding something means its potential is set to zero. This does not imply that the charge becomes zero - it becomes whatever it needs to be for the potential to be zero.

I see, so if I had a charge Q inside and Q outside, when I ground the outer shell the charge would become -Q?

 

[Orodruin]

I see, so if I had a charge Q inside and Q outside, when I ground the outer shell the charge would become -Q?

Yes.

 

[Davidllerenav]

Yes.

Ok. So in this problem, once I ground the outer shell, I would have a new charge q, and the potential at R2 would be the integral from infinity to 0, right? But I'm confused about the field, wouldn't it be 0 since the charge enclosed if r>R2 is Q-Q?

 

[Orodruin]

Ok. So in this problem, once I ground the outer shell, I would have a new charge q, and the potential at R2 would be the integral from infinity to 0, right? But I'm confused about the field, wouldn't it be 0 since the charge enclosed if r>R2 is Q-Q?

Yes. The field outside is zero.

For any region with V = 0 on its boundary (in the case of your spheres and grounding the outer, infinity also counts when we look at the region outside) and no charge in the region, V = 0 inside that region and so the field is zero in that region.

 

[Davidllerenav]

Yes. The field outside is zero.

For any region with V = 0 on its boundary (in the case of your spheres and grounding the outer, infinity also counts when we look at the region outside) and no charge in the region, V = 0 inside that region and so the field is zero in that region.

It is zero, but when I ground it, I can say that the outer shell gets a q' charge, thus, the potential at R2 would be $V(R_2)=-\int_\infty^{R_2} \vec E\cdot\vec l=-\int_\infty^{R_2} \frac{q'+Q}{4\pi\epsilon_0 r^2}dr=\frac{q'+Q}{4\pi\epsilon_0 R_2}$ and since the potential at R2 is zero, I get $0=\frac{q'+Q}{4\pi\epsilon_0 R_2}\Rightarrow q'=-Q$, right? And that's why nothing happens to the charge of the outer shell when it is grounded, as the problem says.

 

[Orodruin]

The shorter explanation would be that the outer shell is at zero potential already in the original setup, but yes. Grounding the shell does not change its potential because it was already at zero potential.

Now how about grounding the inner shell?

 

[Davidllerenav]

The shorter explanation would be that the outer shell is at zero potential already in the original setup, but yes. Grounding the shell does not change its potential because it was already at zero potential.

Now how about grounding the inner shell?

Grounding the iner potential would make the inner potential to have a final charge q1, and integrating as before, I can write thar $V(R_1)=\frac{q_1}{4\pi\epsilon_0 R_1}-\frac{Q}{4\pi\epsilon_0 R_2}=0$ thus $\frac{q_1}{R_1}=\frac{Q}{R_2}\Rightarrow q_1=Q\frac{R_1}{R_2}$, right?

 

[Orodruin]

Grounding the iner potential would make the inner potential to have a final charge q1, and integrating as before, I can write thar $V(R_1)=\frac{q_1}{4\pi\epsilon_0 R_1}-\frac{Q}{4\pi\epsilon_0 R_2}=0$ thus $\frac{q_1}{R_1}=\frac{Q}{R_2}\Rightarrow q_1=Q\frac{R_1}{R_2}$, right?

Correct.