Equipotential surface (test questions)

[vcsharp2003]

Homework Statement

I am trying to answer the following two questions related to equipotential surfaces.

Relevant Equations

None

The answer to the first question should be a sphere since for very large distances the multiple charges will act as a point charge. 1(a) is correct answer.

For the second question, I find it slightly vague. How can equipotential surface be zero, may be it's asking for the potential of equipotential surface. None of the options makes sense since it's not mentioned if the conductor sphere is charged or not. Also inside a conducting sphere, the equipotential region is not a surface but a three dimensional sphere.

If I assume the sphere is charged in question 2, then the best answer would be 2(b) Remains constant.

 

[BvU]

For the second question, I find it slightly vague. How can equipotential surface be zero, may be it's asking for the potential of equipotential surface. None of the options makes sense since it's not mentioned if the conductor sphere is charged or not.

What happens to the charges inside a conductor if there is a potential difference ?

[edit] but I grant you that the wording is an awful 2D/3D mixup..

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[vcsharp2003]

What happens to the charges inside a conductor if there is a potential difference ?

[edit] but I grant you that the wording is an awful 2D/3D mixup..

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Charges would move/flow.

But the question doesn't mention whether the sphere is charged or not. I do understand that for a charged sphere, the potential inside the conductor is the same for all points and is equal to the potential at the surface.

For an uncharged sphere, also no potential difference exists inside it.

Also, is my question 1 answer correct?

 

[BvU]

Charges would move/flow.

until there is no more potential difference $\Rightarrow$ potential is constant. Charged or not.
$\vec E=-\vec \nabla V$ is zero. Potential itself is not necessarily zero. See here.

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[vcsharp2003]

Potential itself is not necessarily zero

Wouldn't it be zero if conducting sphere was uncharged?

 

[BvU]

 

[haruspex]

Wouldn't it be zero if conducting sphere was uncharged?

There is no absolute zero potential. Potentials are always relative to some chosen zero. By convention, it is usually taken to be zero at infinity, but there are exceptions, e.g. field due to infinite uniformly charged plane.

 

[hutchphd]

Homework Statement: I am trying to answer the following two questions related to equipotential surfaces.
Relevant Equations: None

If I assume the sphere is charged in question 2, then the best answer would be 2(b) Remains constant.

There is nothing wrong, in my humble opinion, with the question. If (a) is correct then (b) must also be correct because zero is in fact a constant. Perhaps you were nonplussed by such an inartful statement of the question, but your analysis is correct, and no foul was committed!

 

[BvU]

I have the impression that last line was added later ...

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[vcsharp2003]

I have the impression that last line was added later ...

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Yes.

 

[jbriggs444]

There is nothing wrong, in my humble opinion, with the question. If (a) is correct then (b) must also be correct because zero is in fact a constant. Perhaps you were nonplussed by such an inartful statement of the question, but your analysis is correct, and no foul was committed!

The question is a poor joke.

The question asks: "Equipotential surface inside a conducting sphere is: <blank>". Let us fill in the possible completions in turn:

"Equipotential surface inside a conducting sphere is zero".

No. It is a surface. It is not a number. Nor would it be correct to call it zero in the case of a charged sphere alone in the universe. Nor is it meaningful to give a value at all without having established a reference potential. If one has established a reference at infinity then the charge distribution throughout the rest of the universe becomes relevant. Nobody said that this conducting sphere was alone in an empty universe.

"Equipotential surface inside a conducting sphere is remains constant".

No. This is gibberish. Not even grammatically correct. Presumably the intent is to assert that every equipotential surface that one might imagine within a conducting sphere would share the same potential. Yes, every equipotential surface constructed within an equipotential volume has the same potential as every other. The word "remains" is a poor choice. It suggests that something is changing over time. Nothing is changing here. We have a static sphere and a fixed (albeit unspecified) choice of surface.

"Equipotential surface inside a conducting sphere decreases from center to surface"

No. What does this even mean? What would be decreasing? If they mean the potential of a sequence of surfaces shaped like spherical shells would be decreasing as one follows the shells from center to surface then this is clearly false. Every surface you can draw inside of a conducting sphere is an equipotential surface. All have the same potential. It is an equipotential volume.

"Equipotential surface inside a conducting sphere increases from center to surface"

No. For the same reasons as above. However, if one imagines a family of spheres of varying radii then the surface area of these spheres would indeed increase from center to surface. So this answer is plausibly correct. Nobody said that this question is about potential. It could just as well be about surface area. Surface area is a very reasonable number to associate with a surface.

 

[vcsharp2003]

The question is a poor joke.

The question asks: "Equipotential surface inside a conducting sphere is: <blank>". Let us fill in the possible completions in turn:

"Equipotential surface inside a conducting sphere is zero".

No. It is a surface. It is not a number. Nor would it be correct to call it zero in the case of a charged sphere alone in the universe. Nor is it meaningful to give a value at all without having established a reference potential. If one has established a reference at infinity then the charge distribution throughout the rest of the universe becomes relevant. Nobody said that this conducting sphere was alone in an empty universe.

"Equipotential surface inside a conducting sphere is remains zero".

No. This is gibberish. Not even grammatically correct. If it refers to potential as one wanders over an equipotential surface and if the potential at one point on the surface were zero (very questionable as above) then the statement would be a tautology: all points on an equipotential surface share the same potential by definition. There is nothing special about an equipotential surface inside a conducting sphere in this regard.

"Equipotential surface inside a conducting sphere decreases from center to surface"

No. What does this even mean? What would be decreasing? If they mean the potential of a sequence of surfaces shaped like spherical shells would be decreasing as one follows the shells from center to surface then this is clearly false. Every surface you can draw inside of a conducting sphere is an equipotential surface. All have the same potential. It is an equipotential volume.

"Equipotential surface inside a conducting sphere increases from center to surface"

No. For the same reasons as above.

Exactly, for students who are learning Physics, how can they be expected to understand such a question.

 

[vcsharp2003]

There is no absolute zero potential. Potentials are always relative to some chosen zero. By convention, it is usually taken to be zero at infinity, but there are exceptions, e.g. field due to infinite uniformly charged plane.

But, if a spherical conductor is uncharged and is not in a region of electric field, then we could say that the work done against electrostatic forces in bringing a unit +ve test charge from infinity to a point inside the conductor would be zero since there is no electric field i.e. no electrostatic force against which some work is being done. Therefore, the potential inside the conductor would be zero. I'm not sure if there is something wrong in my above logic.

 

[kuruman]

But, if a spherical conductor is uncharged and is not in a region of electric field, then we could say that the work done against electrostatic forces in bringing a unit +ve test charge from infinity to a point inside the conductor would be zero since there is no electric field i.e. no electrostatic force against which some work is being done. Therefore, the potential inside the conductor would be zero. I'm not sure if there is something wrong in my above logic.

What's wrong with your logic is that the positive test charge will induce charges on the skin of the sphere which will then produce a dipolar field that will exert an attractive force on the test charge. The induced charges are there to ensure that the sphere remains an equipotential but at a higher value of the potential. The force on the test charge would be the same as the Coulomb attraction with its image charge inside the sphere. $$F=-\frac{q^2}{4\pi\epsilon_0}\frac{RD}{\left(D^2-R^2\right)^2}$$ where
$q=~$ test charge
$R=~$radius of the sphere
$D=~$distance of charge from the center of the sphere

Test charges were invented in the approximation that any existing field in their vicinity is much much larger than the field the charge creates, therefore the field created by the test charge can be ignored relative to the existing field. However if there is no field originally, you cannot say that the test charge produces zero field because this implies that it has zero charge. If that were the case, it wouldn't be a test charge, would it?

 

[vcsharp2003]

What's wrong with your logic is that the positive test charge will induce charges on the skin of the sphere which will then produce a dipolar field that will exert an attractive force on the test charge. The induced charges are there to ensure that the sphere remains an equipotential but at a higher value of the potential. The force on the test charge would be the same as the Coulomb attraction with its image charge inside the sphere. $$F=-\frac{q^2}{4\pi\epsilon_0}\frac{RD}{\left(D^2-R^2\right)^2}$$ where
$q=~$ test charge
$R=~$radius of the sphere
$D=~$distance of charge from the center of the sphere

Test charges were invented in the approximation that any existing field in their vicinity is much much larger than the field the charge creates, therefore the field created by the test charge can be ignored relative to the existing field. However if there is no field originally, you cannot say that the test charge produces zero field because this implies that it has zero charge. If that were the case, it wouldn't be a test charge, would it?

Yes, I get it. That's why the test charge is very small so that it doesn't disturb the existing field or the absence of electric field.

I think negative work would be done on the test charge against electrostatic forces (because applied force on test charge and test charge displacement would be in opposite directions). But as the test charge approaches 0, the negative work done would also approach 0. So, in the limiting case, could we say that potential would also approach zero for a point on the surface of uncharged spherical conductor?
($V = \displaystyle{\lim_{q \rightarrow 0^{+}} {\frac {W} {q} }}$)

 

[haruspex]

Therefore, the potential inside the conductor would be zero.

Leaving aside the correction @kuruman makes to the first part of post #13, your conclusion I quote above only follows if you adopt the convention that the potential is zero at infinity.

 

[vcsharp2003]

So, in the limiting case, could we say that potential would also approach zero for a point on the surface of uncharged spherical conductor?
($V = \displaystyle{\lim_{q \rightarrow 0^{+}} {\frac {W} {q} }}$)

I think I'm not correct in my above statement since W will come closer to zero from the -ve side and q would come closer to 0 from the +ve side, but their ratio may not necessarily approach zero (we know from Limits in Calculus that 0/0 is an indeterminate form). So, most likely, the potential at the surface of a neutral conductor would turn out to be negative.

 

[haruspex]

I think I'm not correct in my above statement since W will come closer to zero from the -ve side and q would come closer to 0 from the +ve side, but their ratio may not necessarily approach zero (we know from Limits in Calculus that 0/0 is an indeterminate form). So, most likely, the potential at the surface of a neutral conductor would turn out to be negative.

As the test charge approaches the surface of the sphere, the charge distribution on the sphere will progressively resemble an equal and opposite point charge at the nearest point. The whole will approximate a dipole.

 

[vcsharp2003]

As the test charge approaches the surface of the sphere, the charge distribution on the sphere will progressively resemble an equal and opposite point charge at the nearest point. The whole will approximate a dipole.

Ok.

1. But, would it be correct to say that the potential at a surface point of a neutral conductor would turn out to be slightly negative ( if potential at infinity is taken as 0) using the test charge concept of work done against electrostatic forces?
2. However, ideally speaking, the potential on the surface of a conductor would be 0 if potential at infinity is taken as 0.

I'm not sure if above two conclusions are correct.

 

[haruspex]

the potential on the surface of a conductor would be 0 if potential at infinity is taken as 0.

No, why? If a charge is placed anywhere near an uncharged conductor the potential on the surface of the conductor will be nonzero (relative to that at infinity).
To make its potential zero it would have to be grounded, making it charged.