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Homework Help: Work done by charge in electric field on the xyz plane

  1. Sep 30, 2015 #1
    1. The problem statement, all variables and given/known data

    There is a uniform electric field, E = 270 N/C, parallel to the xz plane, making an angle of 32o with the positive z axis and an angle of 58o with the positive x axis. A particle with charge q= 0.475 C is moved from the point (xi = 2.0 cm, yi = 0, zi = 0) to the point (xf = 8.5 cm, yf = 6.0 cm, zf = -5.5 cm). How much work is done by the electric field.

    2. Relevant equations

    W = qE.d

    3. The attempt at a solution

    The only change in distance I need is parallel to the electric field in the x-z plane.

    I have no idea how to find the distance. This is what I tried.

    d = (0.055m)2+(0.065m)2 = 0.085 m

    Then I believe it's a matter of substituting this into the above equation with the angle which I also don't know how to find. Some guidance would be appreciated thanks.
  2. jcsd
  3. Sep 30, 2015 #2


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    How about expressing the displacement and the field vectors in terms of their components? Remember, you can also write ## \mathbf{A}\cdot\mathbf{B} = A_xB_x + A_yB_y + A_zB_z ##.
  4. Sep 30, 2015 #3
    Could you tell me if this is right?

    E = 270 Cos 58oi + 270 Cos 32ok
    = 143.08 i + 0 j + 228.97 k

    If that's my electric field I basically subtract the distance points to get (d) and put it in W = qE.d to get W. Then I can square the values, add them and square root it to get the magnitude of W?
  5. Sep 30, 2015 #4


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    Your components for ## \mathbf{E} ## look good, but note that ## W ## is a scalar, not a vector. ## W = \Delta \mathbf{r} \cdot q\mathbf{E} ##, and the dot product always yields a scalar.
  6. Sep 30, 2015 #5
    So that means

    Since my ## \Delta \mathbf{r} ## = (0.065 i + 0.06 j -0.055 k)

    Therefore W = | (0.065 i + 0.06 j -0.055 k) ## \cdot ## 0.475 (143.08 i + 0 j + 228.97 k)|
    = |4.4176 - 5.9818|
    = 1.56 J

    Am I correct here? (btw Thanks for your quick responses)
  7. Sep 30, 2015 #6


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    Why take the absolute value? Nothing says that the work has to be positive.
  8. Sep 30, 2015 #7
    Oh yeah good point. So W = -1.56 J. I guess I did it right?
  9. Sep 30, 2015 #8


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    As a check, you could use the fact that ## \mathbf{E}\cdot\Delta\mathbf{r} = E\Delta r\cos\theta ## to get the angle between ## \mathbf{E} ## and ##\Delta \mathbf{r}##. If the angle is more than 90 degrees, you should expect the work to be negative for a positive charge, like in this case. If you throw a rock upward, then while the rock is traveling upward, gravity is performing negative work on it. Same goes for a positive charge moving against the electric field.
  10. Sep 30, 2015 #9
    Ok thanks for your help!
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