Work of Electric Field Problem, can't find mistake

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Homework Help Overview

The discussion revolves around a problem related to calculating the work done by an electric field when assembling multiple charges. The original poster references a specific equation for work and expresses difficulty in completing part D of the problem, which involves the total work to assemble four charges.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use the equation W=qV but is uncertain about how to apply it for part D. They question whether to calculate potential energy for each charge separately and consider the total work involved.

Discussion Status

Some participants have suggested using the formula for electric potential in relation to specific points and indicated that the potential energies might need to be summed based on the charges' signs. There is an indication of progress as the original poster mentions figuring something out, but no explicit consensus or resolution has been reached.

Contextual Notes

The original poster has provided specific values for components A, B, and C, and noted that the work done in assembling two of the charges is zero. There is a reference to external links for additional context, but the completeness of the problem setup remains unclear.

jigs90
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Homework Statement



I posted the question on cramster at
http://answerboard.cramster.com/physics- ...
so you could see the picture.



Homework Equations



W = qV = kq1q2 / r

The Attempt at a Solution



I figure out A, B, and C, (which are the x component, y component, and the electric potential, but I can't figure out how to do part D, (which is the work to asseble the four charges). Please help! I tried to you W=qV, but that didn't work.

My Answers were
A) was 24.4
B) was 77.96
C) was 2.719x10^3

Work done in assembling +6 uF and -6uF is zero

Total work =kq1q2/r=9*10^9*8.2*16.4*10^-12/(47sin(60

Total work =2.97*10^-2 J
 
Last edited by a moderator:
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The link was

http://answerboard.cramster.com/physics-topic-5-189871-0.aspx

sorry
 
Last edited by a moderator:
Wait, should I calculate each PE separately and then add them together? Would there be 6 different PEs?
 
Figured it out...
 
Upon brief inspection, What I would do is use the formula V= kQ/R for each point in relation to P. Then you can add these based on the sign of the charge to solve for W=qV. Give that a try.
 

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