Need clarification on a voltage problem

  • Thread starter Byrgg
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My physics teacher explained this to me, and I'm pretty sure I undertsand, but I want to hear the people here have to say, here's the problem in my textbook:

Suppose the globe of a van de graff generator has a large positive charge. A small metal ball has a positive charge of 1.6 x 10 ^ -8 C, so the globe repels the ball. Assume that as average force of 6.0 x 10 ^ -3 N is needed to move the ball 8.0cm closer to the van de graff generator. How much work is done and what is the change in the electric potential of the ball?

I know how to do this:

W = Fd
W = (6.0 x 10 ^ -3)(0.08m)
W = 4.8 x 10 ^ -4 J

V = W/Q
= (4.8 x 10 ^ -4)/(1.6 x 10 ^ -8)
= 3.0 x 10 ^ 4V

Then there's a couple follow-up questions, Suppose the work done in moving the metal ball between the two points(8.0cm) is 6.5 x 10 ^ -4J, calculate the size of the repulsive force.

This I also did:

W = Fd
F = W/d
= 6.5 x 10 ^ -4/0.08
= 8.1 x 10 ^ -3 N

First off, I'd like to ask, the repulsive force of the van de graff generator must have a reaction force(newton's 3rd law), so then the small metal ball must also exert an equal reaction force on the van de graff generator, right? Is there also an action force exerted on the van de graff generator by small metal ball, and thus a raction force of the van de graff generator on the small metal ball? Or is the only action force the one the generator exerts on the ball?

Second, the second follow-up question asks the following:
suppose the change in electric potential of the charged ball remains 3.0 x 10 ^ 4V. Explain why the electrical potential should stay the same. What is the now the charge on the small metal ball?

I assume they mean to use the work mentioned in the previous problem to get the answer:

V = W/Q
Q = W/V
= (6.5 x 10 ^ -4)/(3.0 x 10 ^ -4)
= approximately the answer in the book, did I do that right?

Anyway, the 'explain' part is the confusing part for me. My teacher said it's because of coulomb's law: F(electrical) = (kQ_1Q_2)/(d ^ 2)

If you substitute that into the voltage equation you get the following:

V = W/Q
= Fd/Q
= (((kQ_1Q_2)/(d ^ 2)) x d)/Q_1

Q_1 cancels out, therefore proving that the charge on the small metal ball(Q_1) has no affect on the voltage. This could explain why the work is increasing, if Q_1 increases, force increases, therefore work increases, but voltage remains the same seeing as it cancels out in the voltage equation. Is this right?

There, a little long, but if someone could clarify my 2 questions(the last paragraph and the paragraph in the middle is where they are found) please? Please respond soon, thanks in advance.
 

Answers and Replies

  • #2
Andrew Mason
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Byrgg said:
First off, I'd like to ask, the repulsive force of the van de graff generator must have a reaction force(newton's 3rd law), so then the small metal ball must also exert an equal reaction force on the van de graff generator, right? Is there also an action force exerted on the van de graff generator by small metal ball, and thus a raction force of the van de graff generator on the small metal ball? Or is the only action force the one the generator exerts on the ball?
You can think of Newton's third law as "all forces come in equal and opposite pairs". There is nothing special about the terms 'action' and 'reaction' and it is probably best to avoid that terminology. Simply put: the force exerted by the charged metal ball on the Van de graff is equal and opposite to the force exerted by the van de graff on the ball.
Second, the second follow-up question asks the following:
suppose the change in electric potential of the charged ball remains 3.0 x 10 ^ 4V. Explain why the electrical potential should stay the same. What is the now the charge on the small metal ball?

....

Q_1 cancels out, therefore proving that the charge on the small metal ball(Q_1) has no affect on the voltage. This could explain why the work is increasing, if Q_1 increases, force increases, therefore work increases, but voltage remains the same seeing as it cancels out in the voltage equation. Is this right?
Since electric potential of the van de graf is "energy per unit charge", it does not depend on the charge of the ball.

AM
 
  • #3
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Andrew Mason said:
You can think of Newton's third law as "all forces come in equal and opposite pairs". There is nothing special about the terms 'action' and 'reaction' and it is probably best to avoid that terminology. Simply put: the force exerted by the charged metal ball on the Van de graff is equal and opposite to the force exerted by the van de graff on the ball.
AM
Ok, I know they come in pairs but is there really 4 forces at work here, or just 2? I know that if the generator exerts a force on the ball, then the ball exerts a force on the generator, but is there another set of forces? A force applied by the ball to the generator which is responded to by the force from the generator? Or is the only applied force the one applied by the generator on the ball(with a force being applied back to it).

Andrew Mason said:
Since electric potential of the van de graf is "energy per unit charge", it does not depend on the charge of the ball.
Isn't the electric potential measured here the one on the ball? The question gives the charge of the ball, and a way to find the work, so I'd assume the question is implying that you find the electric potential of the ball.
 
  • #4
Andrew Mason
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Byrgg said:
Ok, I know they come in pairs but is there really 4 forces at work here, or just 2? I know that if the generator exerts a force on the ball, then the ball exerts a force on the generator, but is there another set of forces? A force applied by the ball to the generator which is responded to by the force from the generator? Or is the only applied force the one applied by the generator on the ball(with a force being applied back to it).
There are various ways to think of the interaction between the two. One way to think of it is an electric field [itex]\vec E[/itex] created by the van de graf. When the ball is placed in that field, it experiences a force [itex]\vec F = q_{ball}\vec E[/itex] and a potential energy equal to the energy required to bring the charge from infinity to r, (r = distance from center of ball to center of van de graf).

Isn't the electric potential measured here the one on the ball? The question gives the charge of the ball, and a way to find the work, so I'd assume the question is implying that you find the electric potential of the ball.
Electric potential is the potential energy per unit charge at a point in an electric field. In this case, the electric field is that of the van de graf. So the potential is a function of position in that field.

If you were asked the potiential with respect to the field of the small ball, you would have to consider the energy per unit charge with respect to the electric field of the small ball.

AM
 
  • #5
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There are various ways to think of the interaction between the two. One way to think of it is an electric field created by the van de graf. When the ball is placed in that field, it experiences a force and a potential energy equal to the energy required to bring the charge from infinity to r, (r = distance from center of ball to center of van de graf).
Yeah, the ball experiences a force applied by the generator, I know this, it is equal to the energy required to move the charge the given distance, right? And therefore the ball has a reaction force to this force, but what I'm asking is if the ball applies it's own(non-reaction force) to the van de graff generator, and thus experiences a reaction force.

Electric potential is the potential energy per unit charge at a point in an electric field. In this case, the electric field is that of the van de graf. So the potential is a function of position in that field.

If you were asked the potiential with respect to the field of the small ball, you would have to consider the energy per unit charge with respect to the electric field of the small ball.
The question clearly says what is the change in the electric potential of the ball for the first part, doesn't that imply that the follow-up questions are the same? Or is this electric potential caused by the electric field of the generator?

Also, the energy required to move the charge on the ball is a result of what? The electric field of the ball, or the generator?
 
  • #6
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Someone please help.
 
  • #7
nrqed
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Byrgg said:
Yeah, the ball experiences a force applied by the generator, I know this, it is equal to the energy required to move the charge the given distance, right?
This is a bit of a confusing statement. Force and energy cannot be said to be equal (I know you know that, I am just pointing out that one must be careful about sayingthis clearly). There is a force, and from the force one can calculate the work done required in moving an object.
And therefore the ball has a reaction force to this force, but what I'm asking is if the ball applies it's own(non-reaction force) to the van de graff generator, and thus experiences a reaction force.
two points:

first, as Andrew said, the terminology "action-reaction" is a bit misleading. It is not that the ball applies a force on the generator and the generator applies a reaction force to the ball *AND* that the generator applies another force to the ball and the ball applies another reaction force to the generator! There is only *one* pair of action-reaction forces acting on the ball and on the generator. Which one you call the action and which one you call the reaction is purely arbitrary.

Second point: strictly speaking , there are no "non-reaction" forces. For every single force an object creates on a second object, there will be a "reaction force" of the second object on the first. Always.

(and again, someone could present things the other way around: they could say that what you call the reaction force of the second object on the first one is actually an *action* of the second object on the first and that what you call the action of the first on the second is actually the *reaction* of the first on the second. It's semantics. Like you saying this object is to the right of that other object and someone facing you saying NO! the first object is to the left of the second one)
 
  • #8
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Okay, so there's not 4 forces here? Just one ball applying a force to the other, while the other providing a reaction force? But each ball has it's own charge, there still aren't 4 forces here? Anything else you could say to clarify this a bit more? And also if you could help answer the last bit there I would be greatful, there's still a chunk of this I don't understand.
 
  • #9
nrqed
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Byrgg said:
Okay, so there's not 4 forces here? Just one ball applying a force to the other, while the other providing a reaction force? But each ball has it's own charge, there still aren't 4 forces here? Anything else you could say to clarify this a bit more? And also if you could help answer the last bit there I would be greatful, there's still a chunk of this I don't understand.
I will get to the other question in a minute.

And yes, there are really only two forces (one on each object). A force exerted by the ball on the generator and one exerted by the generator on th eball. That's it. which is called the action and which is called the reaction force is a matter of point of view. (it's like if you push on the wall and the wall pushes back on you. It is not that you exert a force on the wall and it exerts a reaction force on you AND that the wall exerts a force on you and you exert a reaction force on the wall. There are really only two forces.)
 
  • #10
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Yeah I understand your example there with the wall, but both balls have a charge, that doesn't mean anything? I just thought that the ball having a charge, would mean that it also generates another force...
 
  • #11
nrqed
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Byrgg said:
Yeah I understand your example there with the wall, but both balls have a charge, that doesn't mean anything? I just thought that the ball having a charge, would mean that it also generates another force...
I amnot sure by what you mean by "another" force.

Imagine that the ball was neutral. Then there would be no force at all, right? It would not create any force on the generator and the generator would not create any force on the ball! If the ball is charged, the ball exerts a force on the generator and vice versa. If either the ball or th egenerator is neutral, there is no force at all. You see what I mean?
 
  • #12
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If one is neutral than there's no force at all? I understand the example when you put it that way, but why is that?
 
  • #13
Hootenanny
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Byrgg said:
Yeah I understand your example there with the wall, but both balls have a charge, that doesn't mean anything? I just thought that the ball having a charge, would mean that it also generates another force...
No, it doesn't create another force, but it does contribute to the force acting between them. The force between two point charges is given by;

[tex]F = \frac{Q_{1}Q_{2}}{4\pi\epsilon_{0}r^2}[/tex]

This is the magnitude of the force exerted by the charges on each other. Do you follow?
 
  • #14
Hootenanny
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Byrgg said:
If one is neutral than there's no force at all? I understand the example when you put it that way, but why is that?
Because the electrostatic force only acts between objects with charge. If there is only one charge, what is there to exert a force?
 
  • #15
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Oh ok, that makes sense, so there is only a force when the 2 objects are charged then? And that equation represents the force? Yeahm if one was 0, then the force would be 0, I should've known that already sorry about that.

Now that I think I understand that, could someone explain the remaining part?
 
  • #16
nrqed
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Byrgg said:
If one is neutral than there's no force at all? I understand the example when you put it that way, but why is that?
Do you know the equation for the electric force between two charges? It is [itex] k {q_1 q_2 \over r^2} [/itex] (that's strictly true only for point charges but for any extended object, one may simply use the superposition principle to sum over all the point charges). It involves the *product* of the charges of the two objects. If *either* of the two objects is neutral, there is no electric force on either object.


(again, this is true for true point charges. In real life, some macroscopic objects may have a net charge equal to zero and still attract or repel each other, this is because they really do contain point charges, the eelctrons and protons, and even though there might be an equal number of protons and electrons (the net charge is zero), there might be a net electric force between macroscopic obeject which are neutral but there is no need to get into this here)
 
  • #17
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Byrgg said:
My physics teacher explained this to me, and I'm pretty sure I undertsand, but I want to hear the people here have to say, here's the problem in my textbook:

Suppose the globe of a van de graff generator has a large positive charge. A small metal ball has a positive charge of 1.6 x 10 ^ -8 C, so the globe repels the ball. Assume that as average force of 6.0 x 10 ^ -3 N is needed to move the ball 8.0cm closer to the van de graff generator. How much work is done and what is the change in the electric potential of the ball?
As stated, the question does not make sense. It does not make sense to talk about the change of electric *potential* of an object. One can talk about the change of electric *potential energy* of an object as it moves between two points. OR one can talk about the difference of electric potential between two points in space, in volts (that could be between the initial and final positions where the ball was located). But asking the difference of potential of an object is not defined.
 
  • #18
nrqed
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Byrgg said:
My physics teacher explained this to me, and I'm pretty sure I undertsand, but I want to hear the people here have to say, here's the problem in my textbook:

Suppose the globe of a van de graff generator has a large positive charge. A small metal ball has a positive charge of 1.6 x 10 ^ -8 C, so the globe repels the ball. Assume that as average force of 6.0 x 10 ^ -3 N is needed to move the ball 8.0cm closer to the van de graff generator. How much work is done and what is the change in the electric potential of the ball?

I know how to do this:

W = Fd
W = (6.0 x 10 ^ -3)(0.08m)
W = 4.8 x 10 ^ -4 J

V = W/Q
= (4.8 x 10 ^ -4)/(1.6 x 10 ^ -8)
= 3.0 x 10 ^ 4V

Then there's a couple follow-up questions, Suppose the work done in moving the metal ball between the two points(8.0cm) is 6.5 x 10 ^ -4J, calculate the size of the repulsive force.

This I also did:

W = Fd
F = W/d
= 6.5 x 10 ^ -4/0.08
= 8.1 x 10 ^ -3 N

First off, I'd like to ask, the repulsive force of the van de graff generator must have a reaction force(newton's 3rd law), so then the small metal ball must also exert an equal reaction force on the van de graff generator, right? Is there also an action force exerted on the van de graff generator by small metal ball, and thus a raction force of the van de graff generator on the small metal ball? Or is the only action force the one the generator exerts on the ball?

Second, the second follow-up question asks the following:
suppose the change in electric potential of the charged ball remains 3.0 x 10 ^ 4V. Explain why the electrical potential should stay the same. What is the now the charge on the small metal ball?

I assume they mean to use the work mentioned in the previous problem to get the answer:

V = W/Q
Q = W/V
= (6.5 x 10 ^ -4)/(3.0 x 10 ^ -4)
= approximately the answer in the book, did I do that right?

Anyway, the 'explain' part is the confusing part for me. My teacher said it's because of coulomb's law: F(electrical) = (kQ_1Q_2)/(d ^ 2)

If you substitute that into the voltage equation you get the following:

V = W/Q
= Fd/Q
= (((kQ_1Q_2)/(d ^ 2)) x d)/Q_1

Q_1 cancels out, therefore proving that the charge on the small metal ball(Q_1) has no affect on the voltage. This could explain why the work is increasing, if Q_1 increases, force increases, therefore work increases, but voltage remains the same seeing as it cancels out in the voltage equation. Is this right?
Ok, after reading all the way to the end I realize that what they wanted was really *the difference of potential between the initial and final positions of the ball*. It is very misleading to call this the *difference of potential OF the ball !!A difference of potential is not a property of an object located at a given position it is not a property of the ball.

And to be honest, the whole problem is very very misleading. If the work done has changed, it could be either because the charge on th eball OR the charge on the generator OR both charges have been changed. without specifying which situation it is, the problem is meaningless. It sounds as if they wanted you to consider the case when the charge on the generator was kept fixed.

Your explanation is correct mathematically. But the physical explanation is this: What you found is the potential difference between the initial and final positions of the ball. And this difference of potential is due to all the other charges present in the setup (apart from the ball itself). So it is *only* due to the generator in this problem. If the charge of th egenerator is kept unchanged, then the difference of potential between any two points produced by the generator is left unchaged.

(notice that to calculate the work done to move the ball, one uses only the potential produced by the generator. The ball also produces its own potential so if you would move a third charge around, you would have to include the total electric potential, produced by both the generator and the ball)
 
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  • #19
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Ok, so basically I had the right answer to the last part? The potential difference between the positions of the ball is unrelated to the charge of the ball itself right? As can be seen by the cancelling out of it when you apply the equation for the force between the 2 charges. Does that all sound ok then?
 
  • #20
nrqed
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Byrgg said:
Ok, so basically I had the right answer to the last part? The potential difference between the positions of the ball is unrelated to the charge of the ball itself right? As can be seen by the cancelling out of it when you apply the equation for the force between the 2 charges. Does that all sound ok then?
You got the right idea.
It depends on how rigorous you want to be. You are right about the principle but the derivation is invalid because one cannot use the force equation kq_1q_2/r^2 to represent the *average* force between two points (which is what you are using in your calculations). The real demonstration would involve an integral but you would still find out that the charge of th eball would cancel out, yes.
 

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