Moving Point Charges - Voltage Calculations

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somasimple
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Hi all,

Here is my problem:

Two positive points charges are situated as the initial figure:
1/ How to compute the voltage between A and B (or C)?

2/ If the final condition is like described in the second figure, what is now the voltage between A and B.
3/ if the transition between the two states takes a time t1, does the speed of this transition changes something about the voltage that exists between A and B?

Thanks.
 
Attachments
  • Diagram of two positive point charges q1 and q2 on a horizontal line between points A and B, with point C marked near B, distances r1 (q2 to C) and r2 (q1 to B) and total separation d labeled, captioned initial condition.
    moving_charges_02.jpg
    16.1 KB · Views: 443
  • Diagram of the final condition: vertical line A at left and line B at right separated by distance d, with charges q1 and q2 as spheres near B, and distances r11 from q2 to a point on B and r21 from q1 to B.
    moving_charges_58.jpg
    13 KB · Views: 469
BTW, I received a warning/infraction but I'm not an undergrad student.
The question is just outside my level of expertise (health).
 
somasimple said:
Yes, exactly, that is the way to go.
But since there is a space behind charges I suppose the computation a little different?
No, it's really the same computation. Use the distance from the point of interest (either A or B) to each charge.

somasimple said:
BTW, I received a warning/infraction but I'm not an undergrad student.
The question is just outside my level of expertise (health).
Okay, but that is not relevant. Any textbook-style problem, even if it's for your own independent study, is subject to our "homework help" rules. If you haven't already, you can check out our policy by clicking the "Rules" link at the top of this page, and then scroll down to the section titled Homework Help.
 
Redbelly98 said:
Okay, but that is not relevant. Any textbook-style problem, even if it's for your own independent study, is subject to our "homework help" rules. If you haven't already, you can check out our policy by clicking the "Rules" link at the top of this page, and then scroll down to the section titled Homework Help.
I made the subject. I made the images. I put the subject in the right forum (Classics Physics) but the subject was moved by a moderator.
 
If you'd like to continue with working on the problem, I am willing to help.

As a start, I can suggest computing the potential at point B in your first figure.
 
Redbelly98 said:
As a start, I can suggest computing the potential at point B in your first figure.

My shortcut (?)
k = coulomb's constant
d2 = d-r2
VAB=(k*q1/(r1+d2))+(k*q2/(r2+d2))
it takes account of the space behind the charges (?)
 
Ah, no, it doesn't really work that way. For example, the potential at point B due to charge q1 alone would be

k q1 / (distance from q1 to B)​

However, I am looking at the figures more closely and some things are confusing me:

1. You have used r2 to refer to the position of q1. Did you mean to mix the subscripts like that? It is confusing to do so, but we can go with that definition if that is what you intend.

2. Do "A" and "B" refer to points along the line joining the charges? It would make sense if they do, but then point "C" would be the same as point "B" in your first figure.
 
there is an error in the figure. r1 was meant to belong to q1.
and r2 to q2.
 
Redbelly98 said:
2. Do "A" and "B" refer to points along the line joining the charges? It would make sense if they do, but then point "C" would be the same as point "B" in your first figure.
Yes.
 
Okay, thanks.

So, if you just had charge q1, the potential at B would be
k q1 / r1​
But you also need to add the potential due to charge q2, so you'd have
VB = (k q1 / r1) + ____​

Then you'd do the same thing for VA, the potential at point A.
 
VB = (k q1 / r1) + (k q2 / r2)
VA = (k q1 /(d-r1)) + (k q2 /(d- r2)) ?
 
Yes, that's right.

And the voltage between A & B would be VA-VB ... or VB-VA, depending on whether A or B is being used as the reference point.

EDIT:
Your question #2 works the same way.
#3: As long as we are ignoring relativistic effects, then no, the speed of the transition does not affect the potential.
 
Redbelly98 said:
#3: As long as we are ignoring relativistic effects, then no, the speed of the transition does not affect the potential.
Does it affect the electric current?
 
I thought there was a current since there is a voltage but and I'm certainly wrong.
 
I made new pictures:
Hope they are more accurate.
 
Attachments
  • Diagram of a horizontal line with points A and B at its ends and two charged spheres labelled q1 and q2 between them, with distance d across the top and r1 and r2 marked below and above the line, captioned initial condition.
    moving_charges_02.jpg
    17.2 KB · Views: 454
  • Diagram of a final condition on a straight line: points A and B are a distance d apart, with charges q1 and q2 between them. Distance r12 runs from q1 to B and r21 from q2 to B. Labelled 'final condition'.
    moving_charges_58.jpg
    14.8 KB · Views: 447
somasimple said:
I thought there was a current since there is a voltage but and I'm certainly wrong.
There would have to be some type of conductor joining A and B for there to be a current.
somasimple said:
I made new pictures:
Hope they are more accurate.
Looks good, that more accurately depicts the situation you are describing.
 
Redbelly98 said:
There would have to be some type of conductor joining A and B for there to be a current.
That is what I learned some time ago. A current may exist only if there is an electric circuit.

Redbelly98 said:
Looks good, that more accurately depicts the situation you are describing.
Thanks.