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Electric potential and line of charge; Charge placement between two charges for equilibrium

  • #1

Homework Statement



An electric field is created by an uniform linear charge with a linear charge density is 40 nC/m. Find the potential difference between the two points, one of which is situated 5 cm from the line charge and the second at 10cm.

Two charges, q and -3q, are situated at a distance of 70 cm from each other. At which point will a charge q1 be in equilibrium position in the electric field created by the two above mentioned charges?

Homework Equations



No integrals or derivatives, the teacher said.

E = lambda/2*pi*e0*r

E = V/m

F = kQ1Q2/r**2

The Attempt at a Solution



First one:
[/B]
So, I know that it is a line charge field thus E = lambda/2*pi*e0*r, and that E = V/m so in my logic the voltage of both point 1(5 cm) and point 2(10cm) would be the same, because dividing their electric field by their respective distances would lead to their potential difference being 0. But I know that the potential increases as you get away from the linear charge.

Second One

Should I compare them both with Coulomb's law? I also have some issue understanding what the teacher is actually asking, is he talking about a third charge being put on the field and the point at which all of them would be in equilibrium? My teacher's spoken English is very good, but I find it somewhat troublesome to understand his writing sometimes.
 

Answers and Replies

  • #2
gneill
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Hi againstcalculus,

Welcome to Physics Forums!

In future please place separate questions in separate help requests and choose thread titles that describe the individual questions (so that helpers with the correct skill set can quickly identify the nature of the question in the list of threads). In this case I have changed your thread title to better reflect the nature of the problems.

For the first question I myself am not aware of a non-calculus method to find the potential difference (without just invoking a "canned" formula where someone else has done the calculus to derive it). The electric field is changing nonlinearly as the distance from the wire increases, so really you need to integrate the field along the path from one point to the other.

For the second question you want to find a location where the new charge q1 will remain at rest (no net force acting on it). Assume that the other two charges are somehow fixed in place and cannot move.
 
  • #3
Hi againstcalculus,

Welcome to Physics Forums!

In future please place separate questions in separate help requests and choose thread titles that describe the individual questions (so that helpers with the correct skill set can quickly identify the nature of the question in the list of threads). In this case I have changed your thread title to better reflect the nature of the problems.

For the first question I myself am not aware of a non-calculus method to find the potential difference (without just invoking a "canned" formula where someone else has done the calculus to derive it). The electric field is changing nonlinearly as the distance from the wire increases, so really you need to integrate the field along the path from one point to the other.

For the second question you want to find a location where the new charge q1 will remain at rest (no net force acting on it). Assume that the other two charges are somehow fixed in place and cannot move.
Would it be correct to think that charge q1 would be to the left of q?
 
  • #4
gneill
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Would it be correct to think that charge q1 would be to the left of q?
You'll have to show your reasoning. We can neither confirm nor deny what might be a guess...
 
  • #5
You'll have to show your reasoning. We can neither confirm nor deny what might be a guess...
In order for q1 to be in equilibrium to the other two, both others equilibrium with q1 would have to be equal, and since 5q is 5 times bigger than q and opposite, it's safe to say that it'd have to be to the left of q to be distant enough for equilibrium?

kq1q/(distance )**2 = kq15q/(distance + .70)**2
 
  • #6
haruspex
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E = V/m
No, that is only for a uniform field. More generally, V=-∫E.dr. Although you must not use calculus in your submitted answer, at least you can use that to see what the answer should be.

But I know that the potential increases as you get away from the linear charge.
Really?
 
  • #7
No, that is only for a uniform field. More generally, V=∫E.dr. Although you must not use calculus in your submitted answer, at least you can use that to see what the answer should be.

Really?
But since it has an uniform charge distribution, wouldn't it be an uniform electric field?

I understand that with calculus, after deriving, it would be lambda/2pie0 * ln (0.10 - 0.05), but still can't figure out how to do without it.
 
  • #8
haruspex
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But since it has an uniform charge distribution, wouldn't it be an uniform electric field?

I understand that with calculus, after deriving, it would be lambda/2pie0 * ln (0.10 - 0.05), but still can't figure out how to do without it.
A uniformly charged infinite plate has a uniform field each side, but for a uniform line charge the field diminishes with distance from the wire. The field is only sure to be uniform in the same dimension as the uniformity of the charge. I.e. all points along a line parallel to the charge line will have the same field.
 
  • #9
A uniformly charged infinite plate has a uniform field each side, but for a uniform line charge the field diminishes with distance from the wire. The field is only sure to be uniform in the same dimension as the uniformity of the charge.
Well, so in that case is it implied that it is impossible to solve that without calculus?
 
  • #10
haruspex
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Well, so in that case is it implied that it is impossible to solve that without calculus?
No, there could be a clever way of approaching the question which avoids calculus. However, the presence of ln(2) in the answer implies it will have to be quite clever.
 
  • #11
No, there could be a clever way of approaching the question which avoids calculus. However, the presence of ln(2) in the answer implies it will have to be quite clever.
What would be a starting point?
 
  • #12
haruspex
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What would be a starting point?
The only idea I have had is some sort of scaling argument, but I have not been able to make it work.
As gneill pointed out, there is no way to solve it from first principles without calculus. That would have been needed to obtain the equation you quoted. But maybe there is a standard formula you are expected to know that can be applied here. Consult your course notes.
 
Last edited:
  • #13
gneill
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In order for q1 to be in equilibrium to the other two, both others equilibrium with q1 would have to be equal, and since 5q is 5 times bigger than q and opposite, it's safe to say that it'd have to be to the left of q to be distant enough for equilibrium?

kq1q/(distance )**2 = kq15q/(distance + .70)**2
It sounds like you're on the right track, although the numbers you're using for the second fixed charge seem to keep changing ? The problem statement says -3q, you've just used 5q then 15q in your expression o_O
 
  • #14
haruspex
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It sounds like you're on the right track, although the numbers you're using for the second fixed charge seem to keep changing ? The problem statement says -3q, you've just used 5q then 15q in your expression o_O
I think the kq15q is supposed to mean kq15q. But as you say, it was given as 3, not 5.
 

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