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JenniferSmart1 Group Title

\[\Delta V=V_b-V_a=\frac{\Delta U}{q_0}=-\int_{a}^{b}\vec{E}\cdot d\vec{l}\]

  • one year ago
  • one year ago

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  1. JenniferSmart1 Group Title
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    |dw:1361068864225:dw| Is that what it means?

    • one year ago
  2. JenniferSmart1 Group Title
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    "For a finite displacement form point a to point b, the change in potential is..." (see equation above) I'm trying to make a visual for this concept. I believe my drawing is wrong

    • one year ago
  3. JenniferSmart1 Group Title
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    "The potential difference \(V_b-V_a\) is the negative of the work per unit charge done by the electric field on a test charge when the test charge moves from point a to point b (along any path)."

    • one year ago
  4. TuringTest Group Title
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    |dw:1361069290883:dw|

    • one year ago
  5. TuringTest Group Title
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    Only the component of the motion of the test charge parallel to the E field, i.e. perpendicular to the equipotential surfaces, contribute to the change in potential.

    • one year ago
  6. JenniferSmart1 Group Title
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    Why would E be in the direction from a to b?

    • one year ago
  7. TuringTest Group Title
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    Well, it could be the other way, but usually a is initial and b is final...

    • one year ago
  8. TuringTest Group Title
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    a positive test charge will move from a to b if the field points from a to b

    • one year ago
  9. JenniferSmart1 Group Title
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    If a and b are the initial and final position of the point charge q....ohhhhh Is the particle moving due to the electric field?

    • one year ago
  10. JenniferSmart1 Group Title
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    I see......!!!!|dw:1361069892440:dw|

    • one year ago
  11. JenniferSmart1 Group Title
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    Sorry @TuringTest , I didn't realize that the electric field is causing the particle's displacement. We're starting the Electric Potential chapter next week, and I'm trying to read ahead. Let's say we have a positive point charge. Then the displacement would be as followed....|dw:1361070092302:dw| correct?

    • one year ago
  12. TuringTest Group Title
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    yes, and note that the dot product in the integral means that only the path parallel to the field will change either potential energy or electric potential because of\[\vec E\cdot\vec r=\|\vec E\|\|\vec r\|\cos\theta|dw:1361070442394:dw|\]

    • one year ago
  13. TuringTest Group Title
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    oops because of\[\vec E\cdot\vec r=\|\vec E\|\|\vec r\|\cos\theta\]

    • one year ago
  14. JenniferSmart1 Group Title
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    Sorry I didn't quite get that.... which way does the r vector point? tangent to the path?

    • one year ago
  15. TuringTest Group Title
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    |dw:1361070847504:dw|\(\vec r\) is whatever direction some element of the path \(d\ell\) happens to be pointing in

    • one year ago
  16. TuringTest Group Title
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    so for all the parts of the path that are perpendicular to the field, there is no change in potential|dw:1361070976684:dw|

    • one year ago
  17. JenniferSmart1 Group Title
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    yeah that makes sense. cos 90 is zero....

    • one year ago
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