anonymous
  • anonymous
\[\Delta V=V_b-V_a=\frac{\Delta U}{q_0}=-\int_{a}^{b}\vec{E}\cdot d\vec{l}\]
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1361068864225:dw| Is that what it means?
anonymous
  • anonymous
"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
anonymous
  • anonymous
"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)."

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TuringTest
  • TuringTest
|dw:1361069290883:dw|
TuringTest
  • TuringTest
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.
anonymous
  • anonymous
Why would E be in the direction from a to b?
TuringTest
  • TuringTest
Well, it could be the other way, but usually a is initial and b is final...
TuringTest
  • TuringTest
a positive test charge will move from a to b if the field points from a to b
anonymous
  • anonymous
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?
anonymous
  • anonymous
I see......!!!!|dw:1361069892440:dw|
anonymous
  • anonymous
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?
TuringTest
  • TuringTest
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\[\vec E\cdot\vec r=\|\vec E\|\|\vec r\|\cos\theta|dw:1361070442394:dw|\]
TuringTest
  • TuringTest
oops because of\[\vec E\cdot\vec r=\|\vec E\|\|\vec r\|\cos\theta\]
anonymous
  • anonymous
Sorry I didn't quite get that.... which way does the r vector point? tangent to the path?
TuringTest
  • TuringTest
|dw:1361070847504:dw|\(\vec r\) is whatever direction some element of the path \(d\ell\) happens to be pointing in
TuringTest
  • TuringTest
so for all the parts of the path that are perpendicular to the field, there is no change in potential|dw:1361070976684:dw|
anonymous
  • anonymous
yeah that makes sense. cos 90 is zero....

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