Touque and Work from a Dipole Moment

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Homework Help Overview

The problem involves calculating the work required to rotate a dipole moment in an electric field. The dipole consists of two charges separated by a specific distance and is subjected to a defined electric field strength.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the work using an integral formula but questions the limits of integration when transitioning from parallel to antiparallel orientation. Other participants engage in light-hearted commentary, diverting from the technical discussion.

Discussion Status

The discussion is ongoing, with the original poster expressing confusion over the discrepancy between their calculated work and the expected answer. There is no clear consensus or resolution yet, as participants have not provided direct guidance on the calculation.

Contextual Notes

The original poster mentions receiving the correct answer after multiple incorrect attempts, indicating a potential misunderstanding of the problem setup or the application of the formula.

Cllzzrd
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Problem:
A dipole consists of charges +q and -q separated by 0.62nm . It is in an electric field 2.6×104 C/N.

What is the work required to rotate the dipole from being oriented parallel to the field to being antiparallel to the field?

Equations:
. 90
w=[tex]\int[/tex](p*E*sin(x)
. 0

t=p*E*sin(x)

Work:

The first part asked for the dipole moment, which i found to be 9.9E-29 C*M

So i calculated the torque to be 2.6E-24 N*M, and the second part was the torque at 45 degrees. I found that this was 1.8E-24 N*M.

The last part asked for the work if it was turned from parallel to the electric field to antiparallel.

so I used the formula above and kept getting 1.47E-24 J. The correct answer was 5.2E-24 and I cannot figure out how to get that answer.

Disclaimer: The only reason I know the correct answer is because I got the wrong answer too many times and so it showed me the right answer. I need to figure out how to do this problem.
 
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Cllzzrd said:
The last part asked for the work if it was turned from parallel to the electric field to antiparallel.

so I used the formula above and kept getting 1.47E-24 J. The correct answer was 5.2E-24 and I cannot figure out how to get that answer.
Why does your formula have the integral from 0 to 90? You are going from parallel to antiparallel.
 
More importantly, why are you wearing a touque while working on this? Are you in Quebec?
 
D'oh!
 
Doc Al said:
D'oh!

Let's see how long this takes...
 

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