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More Electric Energy questions

  • #1
(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 teh 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, theres where I get stuck, since that single equation can't be applied to all 3 charges seperately since - don't they all effect one charge?



Any help appreciated.
 

Answers and Replies

  • #2
609
0
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 [tex]qV_{total}[/tex]... find the potential for each charge and add them all up to get [tex]V_{total}[/tex]
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fall behind your class , in this chapter, you are doing potential of point charge, the force fomulas k(q1)(q2) / r^2 is outdated...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
 
  • #3
HallsofIvy
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"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.
 
  • #4
609
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HallsofIvy said:
The potential is that total force divided by the charge q.
That is the electric field, not potential....
 
  • #5
vincentchan said:
work done is [tex]qV_{total}[/tex]... find the potential for each charge and add them all up to get [tex]V_{total}[/tex]
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 theres 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 dont fully understand the work done to a charge, or the Electric Potential Energy. :frown:
 
  • #6
609
0
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
 

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