# Potential difference and a conductor

1. Apr 15, 2010

### snshusat161

1. The problem statement, all variables and given/known data

A solid spherical conductor carries a charge Q. It is surrounded by a concentric uncharged spherical shell. The potential difference between the surface of solid sphere and the shell is V. If a charge of -3Q is given to the shell. Then the new potential difference between the above two points will be -

3. The attempt at a solution

In my view, as both the shell share their center and we know that in conductors we assume that charges are concentrated at the center so charge given to any of the shell increase or decrease the potential of both the shell by the same amount and so there will be no change in potential difference. But I want verification from some knowledgeable or learned person.

2. Apr 15, 2010

### MustangGT

I would think of it this way:
Gauss' Law relates the electric field to the charge enclosed by an imaginary symmetric surface, and the electric field is related to the voltage as E = -del(V) = -dE/dr where the last term is the partial derivative of potential w.r.t. radial distance.

p.s. I can't seem to get the LaTex to display the equations properly.

hope this helps

Last edited: Apr 15, 2010
3. Apr 15, 2010

### collinsmark

You might be correct about the numerical change in potential difference (i.e. 0 change), but I'm not sure if your explanation justifies your answer. If I were you, I would derive an equation that expresses the potential difference between the charged sphere and the shell (even if you leave it in integral form). If the equation is not a function of the outer shell's charge, it shows that charging the shell doesn't affect the answer. Giving an appropriate equation is a lot more convincing than having no equation. http://www.websmileys.com/sm/happy/535.gif

I have a couple of critiques to your explanation.

(1) "we know that in conductors we assume that charges are concentrated at the center"

I beg you not to phrase it that way! :yuck: In conductors, the static (steady-state) charge distribution is always spread out around the conductor's surface. Granted, if the charge distribution is spherically symmetric, the mathematical equations are identical to a point charge, for locations outside the sphere. But the actual charge distribution on the sphere is certainly not concentrated at the center, regardless of the fact that the equations are the same.

(2) "so charge given to any of the shell increase or decrease the potential of both the shell by the same amount and so there will be no change in potential difference."

Even if turns out that putting a charge on the outer shell doesn't change the potential difference, changing the charge on the inner sphere most certainly will! So I can't agree with the above statement.

Also, potential is technically always a potential difference. There is a convention to use the r = infinity as one of the points in the difference, but that is merely a convention. It's understood that even when this convention is used, that the potential is the potential difference with respect to r = infinity.

In this problem, I think the best solution is to not use the r = infinity convention. Instead, derive an equation for the potential difference over the path from the r = outer shell's inner radius (instead of r = infinity) to the inner sphere's radius.

4. Apr 15, 2010

### snshusat161

But while deriving I'm getting following problem.

(1) Charge given to outer sphere will try to induce opposite charge on the inner sphere. So it is hard to find the actual charge on the inner sphere

(2) There may be an electric field setup between both the sphere (if charges are not induced on the inner sphere to cancel it out) and so potential will not be constant on the space between both the sphere.

(3) It is also hard for me to get potential at A because unit charge will need to work against electric field from inner as well as outer sphere to come near. And how much both the sphere contribute toward electric field after rearranging charges due to induction is difficult to predict.

5. Apr 15, 2010

### snshusat161

collinsmark, I think You are right. I've got another question similar to this and my idea ain't working here.

A charge Q is distributed over two concentric hollow spheres of radii r and R (>r) such that the surface densities are equal. find the potential at the common center.

Edit: May be this type of questions are ambiguous and cannot be solved in reality.

6. Apr 15, 2010

### collinsmark

I think all three of these things can be solved by asking yourself, "what is the electric field inside a hollow, spherically symmetric shell?" You can use Gauss' Law to help answer that.

The use that answer for the next question, "what is a spherically symmetrical shell's contribution to the electric field inside the shell, even if something else is inside the shell that has the same spherical symmetry?

7. Apr 15, 2010

### collinsmark

This type of problem can be solved in reality. But this problem actually requires the knowledge of the thickness of the shells, and whether they are conducting or not. If that information is not given, then I guess you are supposed to assume that the hollow spheres are very thin (in which case you can still solve the problem in reality).

In any case, you'll still need to think about what the electric field is inside a hollow sphere. And if there is another sphere inside the hollow sphere (sharing the same center), then the question becomes how does the outer, hollow sphere's charge contribute to the electric field inside the sphere (if it contributes at all). Again, rely on Gauss' Law for the answers to these questions.

[Edit: And once you find the answers to these questions, you will still have to do some integration to find the potential with respect to infinity. Break up the path integral into multiple path integrals, each integrating over different electric field configurations.]

8. Apr 15, 2010

### snshusat161

Gauss's law really makes me mad. Everybody while discussing problem in electrostatic keep on saying "Gauss's Law" "Gauss's Law".

9. Apr 15, 2010

### snshusat161

I know that inside the hollow spherical shell, electric field is zero. And if something with similar symmetry is placed then electric field between them may be zero (not sure) if both have same amount of charge with same sign.

10. Apr 15, 2010

### collinsmark

There ya go!

[Edit: Deleted previous comment, after re-reading your statement. The electric field is not zero in the region between a charged, hollow shell and an inner, charged sphere, even if they have the same charge. Sorry to say this, but Gauss' law will show you why.]

That's because Gauss' Law makes such a good friend! http://www.websmileys.com/sm/happy/042.gif Go ahead and introduce yourself. You won't be disappointed. Gauss' law will be more than happy to get you out of a bind (particularly if there is spherical symmetry involved, and sometimes other types of symmetry too if the wires, cylinders or planes are very long).

Gauss' law states that the electric field is proportional to the charge inside the Gaussian surface (if appropriate symmetry is involved) -- anything outside the Gaussian surface makes no contribution. So inside a completely hollow, charged sphere, there is no charge within the Gaussian surface, so the electric field is zero. If there is something else inside the hollow sphere sharing the same center (with a spherically symmetrical charge distribution), that's the only thing that matters concerning the electric field (for distances in-between the hollow shell and charged thing inside).

Once you know the electric fields, you can integrate over a path to find the potential difference (of course, a negative sign is involved btw).

11. Apr 15, 2010

### snshusat161

but tell me why it is not possible that electric lines of force from the outer sphere enters inside the shell to interfere with the lines of force emerging out from the inner sphere.

12. Apr 15, 2010

### collinsmark

There's an analog of Gauss' law which applies to Newtonian gravitation. I find it useful to think about this, since it also applies in principle to Gauss' law.

Imagine if the Earth was hollow such that all of its mass is concentrated just under the Earth's surface. In other words, imagine that the Earth's mass doesn't change, but just that it's now hollow. Also imagine that it's perfectly spherically symmetric (instead of just approximately).

On the surface of this hollow Earth, you wouldn't be able to tell the difference (well, by gravity that is). From the perspective of gravity, everything would be the same as it is on the actual Earth.

But if you were to go inside the hollow-Earth, you would be totally weightless anywhere inside. You might be inclined to think that as you were close to shell, you would be gravitationally attracted to the portion of the shell nearest to you. But that wouldn't be correct. Yes, you are closer to a certain portion of mass, and that portion of mass has a stronger gravitational pull than any other individual portion of mass of equal size on the shell, but there are so many other portions of mass behind you that the gravitational pull of all the portions all adds up in just such a way that everything cancels out. So anywhere inside this Earth-shell you would feel complete weightlessness -- just as though you were in empty space. The shell itself has zero contribution to the gravitational force.

Now place a small, very dense, and massive sphere at center of this hollow Earth (such that it is fixed in position). You (being inside the hollow-Earth) would gravitate to this small, dense mass. And the equations governing your acceleration to this mass are exactly the same as if just you and the mass were in empty space. The fact that you are in this hollow Earth have no bearing on how you fall toward the mass. Concerning the gravitational force, it's as if the hollow Earth wasn't even there.

In a similar vein, you could calculate the gravitational potential with respect to infinity. The Newtonian equations are nearly identical to the electrostatic equations. And likewise, when performing the respective integrations over the path, you only need to take the mass of the hollow Earth itself into consideration for locations outside the hollow part.

13. Apr 15, 2010

### collinsmark

In a sense they do; it's just that they all cancel each other out (due to the spherical symmetry). Any particular line of force entering the shell is completely canceled out by all the other lines of force entering the hollow part of the shell from the all the other directions.

14. Apr 16, 2010

### snshusat161

so finally, here is the solution given by the coaching center

[PLAIN]http://www.monbattle.com/2.jpg[/CENTER] [Broken]

tell me about $$V_(A)^{'}$$​

Last edited by a moderator: May 4, 2017
15. Apr 16, 2010

### snshusat161

Oh! I think I'm understanding what you wanna tell me. You mean that the electric filed exist in the space between inner and outer sphere is only due to inner sphere. So it doesn't matter how much and how many charge or electrons we give to outer sphere cause it will not contribute even a single percent to electric field from where the unit positive charge need to travel against electric field.

16. Apr 16, 2010

### collinsmark

Yes, that right! http://www.websmileys.com/sm/happy/783.gif

No matter how much you charge up a completely hollow, spherical shell, a unit point charge will float around inside the shell as if the shell wasn't even there.

Note that even though the electric field inside a completely hollow, spherical shell is zero (assuming no other non-spherically symmetric charges are around), it may still have a potential. If the chosen path contains sections outside the shell, the potential inside can still be non-zero, even if the electric field inside is zero.

One last note to differentiate between conducting and non-conducting shells.

o Conducting shells: The electric field inside a completely hollow, conducting shell is always zero, no matter what. If you bring a point charge close to the shell, but still on the outside, the shell's charge distribution will automatically realign itself such that the total electric field inside the shell is still zero. The conducting shell "shields" the inside from any external electric fields.

o Non-conducting shells: The contribution to the total electric field inside of a completely hollow, non-conducting spherical shell (with spherically symmetric charge distribution) is zero, but no shielding takes place. If you were to bring a point charge close to the shell, but still on the outside, there would be an electric field inside the shell (from the point charge), just as if the shell wasn't even there.

Last edited: Apr 16, 2010