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

• Davidllerenav
In summary: Grounding the iner potential would make the inner potential to have a final charge q1, and integrating as before, I can write tharIn summary, when the inner and outer shells are grounded, the potential at the point becomes the potential due to the charge on the inner shell minus the potential due to the charge on the outer shell.
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.

Delta2
Davidllerenav said:
When the outer shell is grouded, its potential goes to zero,
Yes, but what was its potential initially?
Davidllerenav said:
-Q=0
Why?
Davidllerenav said:
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.

haruspex said:
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.

haruspex said:
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.

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.

Orodruin said:
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?

Davidllerenav said:
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.

Orodruin said:
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?

Davidllerenav said:
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.

Orodruin said:
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.

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?

Delta2
Orodruin said:
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?

Delta2
Davidllerenav said:
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.

Davidllerenav

## 1. What is the purpose of grounding a spherical shell?

Grounding a spherical shell helps to dissipate any excess charge that may build up on the surface of the shell. This ensures that the electrical potential on the surface of the shell remains constant and prevents any potential hazards or malfunctions.

## 2. How does grounding affect the electric field inside the shell?

When one of two concentric spherical shells is grounded, the electric field inside the shell becomes zero. This is because the excess charge on the surface of the shell is neutralized by the grounding, resulting in a net charge of zero inside the shell.

## 3. Will grounding one of the shells affect the electric field outside of the shells?

No, grounding one of the shells will not affect the electric field outside of the shells. This is because the electric field outside of the shells is determined by the total charge of both shells, and grounding one of the shells does not change this total charge.

## 4. What happens to the charges on the surface of the grounded shell?

When a spherical shell is grounded, the excess charges on the surface of the shell will flow into the ground, leaving the surface with a net charge of zero. This neutralization of charges helps to maintain a constant electrical potential on the surface of the shell.

## 5. Can you ground only one of the concentric spherical shells?

Yes, it is possible to ground only one of the concentric spherical shells. This will result in a zero electric field inside the grounded shell, while the other shell will still have a non-zero electric field. This can be useful in certain situations where a specific electric field is desired inside the shell.

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