# Electric charges and force (tricky test question)

1. Feb 18, 2006

### vmc303

On my physics test yesterday, one of the questions was the following:

At a particular point in space, a charge Q experiences no force. It follows that

A. if charges are nearby, the total positive charge must equal the total negative charge.

B. there is not enough information to tell whether there are nearby charges or not.

C. if charges are nearby, they have the opposite sign of Q

D. there are no charges nearby

E. if charges are nearby, their total charge must add up to zero.

I answered D, but according to the test key, the correct answer was B. I had initially put B, but when I realized that the problem said "no force" instead of "no net force," it was clear that D was the correct answer. Do I have a case? I'm guessing that the teacher meant it to read "no net force," but the distinction is a very important one for this question.

2. Feb 18, 2006

### Staff: Mentor

I don't think you have a case. The electrostatic force experienced by a charge depends on the field it is exposed to. While you can use superposition to add the contributions to the field from all charges, it is inaccurate to say that a charge experiencing no electrostatic force is really being pulled equally in all directions. The charge experiences no force.

3. Feb 18, 2006

### gulsen

I guess "experiences no force" means "(feels) no net force".

4. Feb 18, 2006

### nrqed

They meant no net force. Then, either there is no force at all or there are 2 or more forces cancelling each other. It is therefore impossible to say anything about the charges nearby.

Pat

5. Feb 18, 2006

### vmc303

The problem is that it didn't say "no net force," it said "no force." It's not at all a trivial distinction in this case. The teacher has often harped on the difference between net force and force, and I assumed this was his attempt to trip people up on that subtlety. Given that the charge experience *no* force at all, the only answer choice which logically follows is D. After all, if it experiences no forces, how could there possibly be any charges nearby?

Last edited: Feb 18, 2006
6. Feb 18, 2006

### nrqed

I understand. And you are right, indeed.

But it would be very artificial to say that one knows that an object experiences no force. Physically, this would make no sense because there is no physical way to tell that an object experiences no forces at all. It is impossible, *physically*, to make any statement about anything else than the net force on an object. If you observe a particle in an experiment and it has no acceleration, you will never be able to say anything else than that the net force is zero.

I understand your point of view, and yes, if we talk about semantics, syaing there is no force on the particle would imply that there is no other charge (or any other mass!) around. But from a physics point of view, one can only talk about net force.

This all boils down to how finicky the prof is and if he makes distinction between semantics and physics. I know that when I teach my students, whenever I talk about the force on a particle (with no other specification), it always means the net force. The only exception is if I ask what is the force on that charge due to this other specific charge.

regards

Pat

7. Feb 18, 2006

### vmc303

Interesting. Well, I'll take it to him and see what he says. I'm doubtful the outcome will be in my favor though. The annoying thing is that the teacher has often stressed the difference between force and net force in his lectures, asking questions like "An object is at rest on the top of a table. Is it experience any forces?" If I hadn't remembered that point of his, I would've answered B without hesitation and left it.

8. Feb 18, 2006

### nrqed

Ok. I understand your point. You should discuss this with him (he might give you partial credit :-) ).

But then I would say that the wording of the question is crucial. If it had been written as ''the particle experiences no FORCES'', then I would say that this might be interpreted as meaning no forces at all. But if force is singular, ''the particle experiences no FORCE'', then I personally would always interpret this as meaning no net force.

Good luck!

9. Feb 18, 2006

### Staff: Mentor

But in that example there are multiple forces that really act on the object! It would be incorrect to say that the object "experiences" no force; those forces exert a real physical stress on the object. The charge, on the other hand, "feels" no force whatsoever.

(But I agree that this question is unnecessarily tricky and I wish you luck in getting partial credit. )

10. Feb 19, 2006

### vmc303

I still don't see the relevant difference; or if there is one, it's certainly something that was never taught in lecture, and can't be something he could expect us to know.

How is the force exerted by a charge any different at all than the gravitational force exerted by a mass? And if there are two equal and opposite charges, canceling each other out perfectly, that seems entirely analogous to forces acting on a book on a table.

Also, check out this webpage: http://www.tjhsst.edu/~jdell/PROBSOLS/ch22.pdf?page=5 It's apparently from the publishers of the original problem, and confirms that the answer to the question, in that particular formulation, is the one I gave. That should be my trump card.

Last edited: Feb 19, 2006
11. Feb 19, 2006

### lightgrav

The Electric force exerted on a charge in an E-field is similar to the
gravitational force exerted on a mass ... it is caused by the local Field.
At any point, there is only ONE Electric field; and only ONE gravitational field.

If your class spent effort adding E-field contributions due to charges, as I expect, then your "case" depends on the prof's wording in such examples.
My guess is that your textbook starts with Electric Force, before E-field ...
so it leads students (and encourages teachers) to mistakenly add or subtract Electric Forces.
Good Luck.

12. Feb 19, 2006

### vmc303

The way it was explained was that an electric field at a point is just the superposition of all the charges acting on that point. So, using Coulomb's law, we would just add up the invidual forces to arrive at the E field at that point.

Here are a few illustrative quotes from the Power Point slides:

"The force between two charges is not
affected by the presence of other charges,
even if those charges are ''in the way.'"

"If there are more than two charges, the net
force on any of them is the sum of the
Coulomb forces exerted by each of the other
charges."

Last edited: Feb 19, 2006
13. Feb 19, 2006

### lightgrav

hmm ... he's using "Coulomb force" as shorthand for ficticious 2-particle contribution to "real" net Force. It is typical. It is misleading.
If he won't give you full credit (he does have an "out"),
make him promise to always teach Field-First in future classes.
That way every student will know to add the E-field contributions first,
so he can replace this question with something more important.

14. Feb 19, 2006

### Staff: Mentor

It's not! I was going to use that very example. If you had a mass (m) in the middle between two equally massive bodies, the gravitational force would be zero. The gravitational force depends on the gravitational field, just like the Coulomb force depends on the Coulomb field.
But it's not. The gravitational field acting on the book is not zero, so there's a gravitational force acting on it. And the normal force of the table on the book is a real force as well. So forces do act on the book, even though they add to zero. (A silly example to make the point: The net force on your foot is zero, say. Now have an elephant step on your foot. The net force is still zero, but you would probably not be tempted to say there are no forces acting on your foot.)

This could be your lucky day! (This proves that the problem was not intended to be a subtle test of your understanding of fields, but was just sloppily worded. And since it's a "book" problem, not one that your instructor made up, you have a good shot at getting full credit.)

Again, I would never ask this kind of question unless I had properly taught the material and stressed this point.

I agree with lightgrav.
Exactly!

15. Feb 21, 2006

### vmc303

So I emailed my professor, laying out my case and including the link to that solution set with the problem in it. I didn't get the response I was hoping for. He essentially brushed me off, and told me that learning how to interpret problems and "read between the lines" was an important part of a science education. I was tempted to reply that precise communication is of the utmost importance in science, and that one would expect the teacher to formulate test problems in such a way so that "reading between the lines" was not neccessary. He didn't discuss any of the various physics-related reasons given in this thread for why I could be incorrect; instead, he just said the the problem becomes "trivial" under my interpretation of it, so I should've known it was incorrect.

Is this worth pursuing further? I believe I have a strong case, especially after finding the solution set showing my answer to be correct under that particular ambiguous wording. I also disagree that the problem becomes any more trivial under my interpretation. However, I'm not sure there's much chance the teacher will give any ground on this.

16. Feb 21, 2006

### nrqed

well, what percentage of the semester is that question worth? (not the percentage of that test, the percentage of the entire semester!). That will give you some perspective about whether it's worth it!!

Pat