Two charges, Electrical Potential

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Homework Help Overview

The discussion revolves around the calculation of electric potential and work done in a system of three charges, specifically focusing on two fixed charges and a third charge brought from infinity. The subject area includes concepts of electric potential, work, and potential energy in electrostatics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the correct application of the formula for electric potential and question the distance calculations involved. There is discussion about the implications of setting V=0 at infinity and how it affects the calculations. Some participants also seek clarification on the relationship between work and energy in this context.

Discussion Status

Participants are actively engaging with the problem, with some providing insights into the correct approach for calculating work and potential energy. There is recognition of the need to clarify the distinction between work done by the electric force and work done by the person moving the charge. Multiple interpretations of the problem setup are being explored.

Contextual Notes

There are mentions of potential confusion regarding the conversion of units from centimeters to meters and the correct interpretation of the geometry involved in the problem. Participants are also addressing the need to accurately read and interpret the provided diagram.

rlc
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Homework Statement


upload_2015-1-22_13-57-41.png

Two charges q = 9.0μ C are fixed in space a distance d = 1.5 cm apart, as shown in the figure.

a) With V = 0 at infinity, what is the electric potential at point C?
b) You bring a third charge q = 9.0μC from infinity to C. How much work must you do?
c) What is the potential energy U of the three-charge configuration when the third charge is in place?

Homework Equations


electrical potential=kq/r
μ=E-6
q=9E9

The Attempt at a Solution


I thought that I should find the distance between q and C using the pythagorean theorem.
((9E9)(9.0E-6)/(2.1213 cm))+((9E9)(9.0E-6)/(2.1213 cm))=7.64E4 V
But this didn't work. The online homework said it was wrong.
Is that the right equation to use? How does the V=0 at infinity affect this problem?
 
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What's the SI unit for distance?
 
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meters
Was I right to calculate the distance from q to C? Is converting cm to m the only mistake?
 
rlc said:
meters
Was I right to calculate the distance from q to C? Is converting cm to m the only mistake?
Yes. Yes No - See post below.
 
Last edited:
Note that the legs of the triangle are d/2, not d.
 
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Wow, I really need to learn to read these diagrams!
That worked, thank you!

Part b asks: You bring a third charge q = 9.0μC from infinity to C. How much work must you do?
How do I start with this problem?
 
Work is closely related to energy.
 
The work done by the electric field in moving an electric charge from infinity to point r is
given by:
W= - deltaU= - q(deltaV)= - q(Vr-Vinfinity)= - qVr

I found this equation online. Does it look like the right one to try?
 
That's close. One of the annoying things about this topic is that when you calculate work, you have to be clear whether you're talking about the work done by the person moving the charge or the work done by the electric force. The force exerted by the person counteracts the electric force, so the work done by the person has opposite sign of the work done by the electric force. So, you need to consider whether the formula you are using is for the work done by the electric force or for the work done by the person.
 
  • #10
I'm hoping its work done by electric force :)
 
  • #11
rlc said:
I'm hoping its work done by electric force :)
Yes, the formula you wrote is for the work done by the electric force. So, you have to modify it to get the work that you do in moving the charge to C.
 
  • #12
Alright, so I figured it out!
Work=qV
where V is the number I just calculated for the first part of this question and q is the 3rd charge given
So, Work=(9.0E-6)(1.53E7V)=1.37E2 J

Then for part c, it's asking for the potential energy of the three charge configuration when the 3rd charge is in place, which the equation for is:
(work done)+(kq^2)/d
(1.37E2 J) + (9E9)(9.0E-6)^2/(0.015 m) = 1.86E2 J

Thank you for helping me thru this and finding my mistakes! I really appreciate it!
 

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