Calculating Work to Move a Charge in an Equilateral Triangle

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Homework Help Overview

The problem involves calculating the work required to move a charge in the context of electrostatics, specifically with two fixed charges at the vertices of an equilateral triangle. The charges are 4e and -4e, and the task is to determine the work needed to move a charge q from one vertex to the midpoint of the line joining the fixed charges.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the potential energy and work equations, questioning how to apply them given the different charges involved. There is uncertainty about translating known formulas for equal charges to this scenario.

Discussion Status

Participants are actively engaging with the problem, raising questions about the correct application of formulas and the nature of the work being calculated. There is a mix of attempts to clarify the potential energy at the starting point and the work done by different forces, indicating a productive exploration of the concepts involved.

Contextual Notes

Some participants express confusion about the setup and the appropriate use of constants and variables in their calculations. There is also mention of the need to clarify whether the work is done by an external force or the electric force.

rgold
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Homework Statement


two particles with charges 4e and -4e are fixed at the vertices of an equilateral triangle with sides of length a. If k=1/4 pi Ԑ what quantity of work is required to move a particle with a charge q from the other vertex to the center of the line joining the fixed charges?

Homework Equations


W = -delta U

The Attempt at a Solution


-(4kQq / a - 2kQq / a) = -2kQq / a => 2kQq / a
I know this is for when the charges are equal (both are Q) but I am not sure how to translate that into my problem...
 
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would it be (( 4*1/4 pi Ԑ *4) - (2*1/4 pi Ԑ *-4))/a?
 
rgold said:
I know this is for when the charges are equal (both are Q) but I am not sure how to translate that into my problem...
No, it's for one charge of q and one of Q.
rgold said:
would it be (( 4*1/4 pi Ԑ *4) - (2*1/4 pi Ԑ *-4))/a?
No. Please post your reasoning and working.
 
haruspex said:
No, it's for one charge of q and one of Q.

No. Please post your reasoning and working.
I think I am just getting more confused and that I should start from the beginning... What is the best way to go about this problem?
 
rgold said:
I think I am just getting more confused and that I should start from the beginning... What is the best way to go about this problem?
What is the potential where the charge starts?
 
haruspex said:
What is the potential where the charge starts?
kQq/r
 
rgold said:
kQq/r
And the potential energy, at the time it starts?

You should also be using the brackets appropriately - 1/4πε0 is read as (1/4)*πε0.
 
rgold said:
W = -delta U
Also, which work is this?
 
Qwertywerty said:
Also, which work is this?
Wouldn't that be the work that I'm looking for?
 
  • #10
Qwertywerty said:
And the potential energy, at the time it starts?

You should also be using the brackets appropriately - 1/4πε0 is read as (1/4)*πε0.
So would it be ((4*1/4πε)-(-4*1/4πε))/a?
 
  • #11
rgold said:
Wouldn't that be the work that I'm looking for?
Is that the work done by the external force, or the electric force?
rgold said:
So would it be ((4*1/4πε)-(-4*1/4πε))/a?
As hurspex pointed out, that is wrong. I was merely requesting you to write in a more, correct fashion. You will need to show your working.
 
  • #12
Qwertywerty said:
Is that the work done by the external force, or the electric force?

As hurspex pointed out, that is wrong. I was merely requesting you to write in a more, correct fashion. You will need to show your working.
at first i thought i should be using (1/(4*πε0))*((q1q12)/r) is this correct? and should my answer have an 'a' is it?
 
  • #13
Qwertywerty said:
Is that the work done by the external force, or the electric force?

As hurspex pointed out, that is wrong. I was merely requesting you to write in a more, correct fashion. You will need to show your working.
or do i need to look at it as a dipole moment?
 
  • #14
rgold said:
kQq/r
I meant the total potential, due to the two fixed charges.
 
  • #15
rgold said:
or do i need to look at it as a dipole moment?
You do not care about forces. This is just about potentials.
 
  • #16
rgold said:
or do i need to look at it as a dipole moment?
Can you consider a +4e and -4e as a dipole for any distance from the dipole? What is their constraint?
 

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