Electric work and potential difference

  • Thread starter jg95ae
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  • #1
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I need a little help with this question:

Calculate the work needed to move a 1.5uC test charge from point B to point D. (I have attached the diagram)

The first part of the question required me to find the electric field at the points A, B, C and D. Which I think I figured out ok.
I found Eb = 7.66 x 10^6 and Ed = 0 N/C

So I think the equation that I need to use is:

W = Fd
W = qE(dx)

whereq = 1.5 x 10^-6C and dx = 0.25m

I'm not sure how to figure out E in this case?
 

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Answers and Replies

  • #2
Doc Al
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You could certainly use that approach (calculating the work directly), but since the field is not constant you would have to use calculus to integrate E(dx) over the displacement.

Much easier is to consider the change in electric potential energy as the charge is moved.
 
  • #3
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If I were to use the change in electric potential energy I would use:

change in PE = q x dV so PE = q(Vb-Vd) ?
 
  • #4
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jg95ae said:
If I were to use the change in electric potential energy I would use:

change in PE = q x dV so PE = q(Vb-Vd) ?

Yes, this is correct, although as per convention, to find work done I would use PE = q(Vd-Vb)
 
  • #5
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Just one more thing, my answer seems a little large.

To calculate Vd I add Vd1 + Vd2, and the same for Vb

which works out to be 2kQ/r where Q is the charges on either side? and r is the distance from each charge to point D?
 
  • #6
Doc Al
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Yes, the general form of the potential at D or B will be V = 2kQ/r, where r is the distance from D or B to one of the charges.
 
  • #7
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When I was putting all the numbers in I realized that there is no consideration for the distance between point B and D. Is the work needed to move the charge independent of the distance moved?
 
  • #8
Doc Al
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Any distance dependence is already included in the potential, so all you have to worry about is the potential at each point.
 

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