Potential of a conducting sphere in a conducting shell

In summary: This makes sense now. In summary, a conducting sphere, charged with Q and radius R, is placed inside a conducting shell with charge 2Q and radius 3R. The electric field and potential are calculated for various regions, and the energy needed to construct this system is found to be 14πk2Q2/R. When the two conductive bodies are connected with a wire, the potential on the inner sphere becomes zero, and the system reaches a lower energy state.
  • #1
razidan
75
1

Homework Statement


A conducting sphere, radius R, charged with Q is inside a conducting shell (2R<r<3R) with charge 2Q. Find the electric potential and the energy.

Homework Equations


[itex] \Phi =-\int_{r_1}^{r_2} \vec{E}\cdot\vec{dl}[/itex]
[itex]U=\int_{V}E^2dV[/itex]

The Attempt at a Solution


I think i got it right, and I'm mostly looking for confirmation:
I started with calculating the field everywhere:
[itex]
\vec{E} (r) =
\begin{cases}
0 & \quad \text{if } \text{ r<R}\\
\frac{kQ}{r^2} \hat{r} & \quad \text{if } \text{ R<r<2R}\\
0 & \quad \text{if } \text{ 2R<r<3R}\\
\frac{3kQ}{r^2} \hat{r} & \quad \text{if } \text{ r>3R}\\
\end{cases}
[/itex]

this leads to:
[itex]
\Phi (r) =
\begin{cases}
\frac{kQ}{2r} & \quad \text{if } \text{ r<R}\\
\frac{3kQ}{3R}+\frac{kQ}{2R}-\frac{kQ}{r} \hat{r} & \quad \text{if } \text{ R<r<2R}\\
\frac{3kQ}{3R} & \quad \text{if } \text{ 2R<r<3R}\\
\frac{3kQ}{r} \hat{r} & \quad \text{if } \text{ r>3R}\\
\end{cases}
[/itex]
[itex] \text{and the energy is} \quad 14\pi\frac{k^2Q^2}{R} [/itex]

Thanks,
R
 
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  • #2
There are a couple things that I'm unclear about. What do you mean when you say 'energy'? Do you mean the potential energy of a particle far away from the sphere? This seems a bit unclear. Also, I'm assuming for the first part that you are measuring the potential difference from infinitely far away; there is no real meaning to an absolute electric potential, it all depends on where you take it in reference to. Is there any more information that came with the problem?

I don't know if this was a typo, but when you list the potential energy for r<R you list it as varying with radius r. That shouldn't be the case; the potential within a conducting body should always be constant. A conducting volume is a large equipotential region; the way you have it listed would give an electric field within the sphere of kQln(2r), which is obviously not true.
 
  • #3
John Morrell said:
There are a couple things that I'm unclear about. What do you mean when you say 'energy'? Do you mean the potential energy of a particle far away from the sphere? This seems a bit unclear. Also, I'm assuming for the first part that you are measuring the potential difference from infinitely far away; there is no real meaning to an absolute electric potential, it all depends on where you take it in reference to. Is there any more information that came with the problem?

I don't know if this was a typo, but when you list the potential energy for r<R you list it as varying with radius r. That shouldn't be the case; the potential within a conducting body should always be constant. A conducting volume is a large equipotential region; the way you have it listed would give an electric field within the sphere of kQln(2r), which is obviously not true.

Hello and thanks for the response:
what I mean by energy is the electrostatics energy needed to construct such a system, by bringing infinitesimal amounts of charge, bit by bit from infinity.
and about "r" - it is indeed a typo... for some reason i cannot edit that post.

R
 
  • #4
I agree with your fields but not your potentials. How did you arrive at them?
 
  • #5
haruspex said:
I agree with your fields but not your potentials. How did you arrive at them?

Thanks, missed a "-" sign:
[itex]
\Phi (r) =
\begin{cases}
\frac{3kQ}{2R} & \quad \text{if } \text{ r<R}\\
\frac{3kQ}{3R}-\frac{kQ}{2R}+\frac{kQ}{r} & \quad \text{if } \text{ R<r<2R}\\
\frac{3kQ}{3R} & \quad \text{if } \text{ 2R<r<3R}\\
\frac{3kQ}{r} & \quad \text{if } \text{ r>3R}\\
\end{cases}
[/itex]
 
  • #6
A follow up question (given that the potential in my last response is correct):
The sphere and the thick shell are now connected with a conductive wire. what is the potential and energy now?
This part has me stumped. I know that now the sphere and the shell have the same potential because charged can transfer.
But I'm not sure what to do.

Thanks,
R
 
  • #7
razidan said:
given that the potential in my last response is correct
It was. Sorry, I omitted to confirm.
razidan said:
the sphere and the shell have the same potential because charged can transfer.
Right. What is conserved? What equations can you write relating the charges and the potentials?
 
  • #8
haruspex said:
It was. Sorry, I omitted to confirm.

Right. What is conserved? What equations can you write relating the charges and the potentials?
I understand that the charge is being conserved.
[itex]Q_1 +Q_2 =3Q[/itex]
I'm not sure that writing this is correct, though:
[itex] \Phi(r<R)=\Phi(2R<r<3R) => \frac{3kQ_1}{2R} = \frac{3kQ_2}{3R} [/itex]

I'm not sure if this is correct because the numerical factors all came from the integration of the fields, before the charge was transferred...
 
  • #9
razidan said:
not sure if this is correct because the numerical factors all came from the integration of the fields, before the charge was transferred...
Quite so, you cannot use those. So recalculate the fields and potentials in terms of two unknown charges.
 
  • #10
haruspex said:
Quite so, you cannot use those. So recalculate the fields and potentials in terms of two unknown charges.
Ok, I will try this, Thanks!
 
  • #11
haruspex said:
Quite so, you cannot use those. So recalculate the fields and potentials in terms of two unknown charges.
I have found that the charge on the inner sphere is now zero. Is that correct? If so, Is is because this configuration has a lower energy associated with it, and so the system "prefers" this configuration?
 
  • #12
razidan said:
I have found that the charge on the inner sphere is now zero. Is that correct? If so, Is is because this configuration has a lower energy associated with it, and so the system "prefers" this configuration?
Yes.
 
  • #13
haruspex said:
Yes.
Thank you very much!
 

1. What is the definition of potential in a conducting sphere in a conducting shell?

The potential of a conducting sphere in a conducting shell is the measure of the electric potential energy per unit charge at any point inside or outside the sphere-shell system.

2. How is the potential calculated for a conducting sphere in a conducting shell?

The potential for a conducting sphere in a conducting shell can be calculated using the formula V = kQ/r, where V is the potential, k is the Coulomb's constant, Q is the charge on the sphere, and r is the distance from the center of the sphere to the point of interest.

3. What is the role of the conducting shell in determining the potential of the sphere?

The conducting shell serves as a boundary condition for the potential of the sphere. It ensures that the potential on the surface of the sphere is constant and equal to the potential of the shell, and helps to distribute charges on the surface of the sphere.

4. How does the potential change as the distance from the center of the sphere increases?

The potential decreases as the distance from the center of the sphere increases, following an inverse-square law. This means that the potential decreases as the distance increases, and the rate of decrease is proportional to the square of the distance.

5. Can the potential of a conducting sphere in a conducting shell ever be negative?

No, the potential of a conducting sphere in a conducting shell can never be negative. This is because the potential is a measure of electric potential energy, and the energy of a system cannot be negative.

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