How to Calculate Electric Potential and Work for Multiple Charges?

Click For Summary

Homework Help Overview

The discussion revolves around calculating electric potential and work for multiple charges, specifically focusing on scenarios involving two and three point charges. Participants are exploring the relationships between electric force, potential energy, and work done in these contexts.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to apply formulas for electric force and potential but express uncertainty about how to proceed after calculating forces. Questions arise regarding the application of these formulas to multiple charges and the distinction between electric potential and electric field.

Discussion Status

Some participants have provided guidance on calculating electric potential by suggesting the addition of individual potentials from each charge. Others are questioning the validity of certain approaches and clarifying concepts related to work done and electric potential energy. There is an ongoing exploration of different interpretations and methods without a clear consensus.

Contextual Notes

Participants note potential gaps in understanding related to the concepts of electric potential and work, with some expressing concerns about falling behind in their studies. There are indications of confusion regarding the application of formulas in multi-charge scenarios.

thursdaytbs
Messages
53
Reaction score
0
(1)Two charges of +2.60 x 10^-8 and -5.50 x 10^-8 C are separated by 1.40 m. What is the electric potential midway between them?

The way to solve this, i tried is by saying F = k(q1)(q2) / r^2. So, the force between the two charges is (9x10^9)(2.6x10^-8)(5.5x10^-8) / (1.4^2). Although, from there I'm not sure where to go because that just solves for the force inbetween the two charges and not the EPE or even more specifically the EPE midway between them.


(2) Three charges are located at the corners of a square whose sides are 2.0m in length. The charges are +2.0, +14, and +5.0 microCouloumbs. The empty corner of the square is opposite the +14 micro coloumb charge. How much work is required to bring up a fourth charge of +8microColoumbs and place it at the empty corner?

I started by saying that Electric Force = k(q1)(q2) / r^2,
Then I figured I would apply that to all 3 charges with the +8charge as q2. Next I would add up all the three forces together to find the total amount of force needed. Except, there's where I get stuck, since that single equation can't be applied to all 3 charges separately since - don't they all effect one charge?



Any help appreciated.
 
Physics news on Phys.org
the formulas of eletric potential is V=kq/r, and potential is a form of energy, if you have two sourse of potential, just find the individual V and add them up...
work done is qV_{total}... find the potential for each charge and add them all up to get V_{total}
I think you have already fall behind your class , in this chapter, you are doing potential of point charge, the force fomulas k(q1)(q2) / r^2 is outdated... do some reading b4 posting next time
 
"The way to solve this, i tried is by saying F = k(q1)(q2) / r^2. So, the force between the two charges is (9x10^9)(2.6x10^-8)(5.5x10^-8) / (1.4^2)."
This is the force each exerts on the other. The problem asked for the electric potential half way between them. Imagine a "test" charge q at distance 0.7 m from each charge. What is the force on that test charge due to each (be careful about the directions). What is the total force on that test charge? The potential is that total force divided by the charge q.
 
HallsofIvy said:
The potential is that total force divided by the charge q.

That is the electric field, not potential...
 
vincentchan said:
work done is qV_{total}... find the potential for each charge and add them all up to get V_{total}

I thought W = q(Vb - Va)? Or is that only valid when it's one charge moving from one place to another, and W = qVtotal, when there's more than one charge?

And yeah, I think I've fallen behind because I can fully grasp the idea of F=k(q1)(q2) / r^2, and E = F/q, but I don't fully understand the work done to a charge, or the Electric Potential Energy. :frown:
 
yes, you are right, However, if the charge is came from infinitely far away, Va = kq/r, Va goes to zero as r goes to infinite, W=qV works perfectly fine in your problem (2), hope this answer your question
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
13
Views
2K