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

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|dw:1361068864225:dw| Is that what it means?
"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
"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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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.
Why would E be in the direction from a to b?
Well, it could be the other way, but usually a is initial and b is final...
a positive test charge will move from a to b if the field points from a to b
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?
I see......!!!!|dw:1361069892440:dw|
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?
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\[\vec E\cdot\vec r=\|\vec E\|\|\vec r\|\cos\theta|dw:1361070442394:dw|\]
oops because of\[\vec E\cdot\vec r=\|\vec E\|\|\vec r\|\cos\theta\]
Sorry I didn't quite get that.... which way does the r vector point? tangent to the path?
|dw:1361070847504:dw|\(\vec r\) is whatever direction some element of the path \(d\ell\) happens to be pointing in
so for all the parts of the path that are perpendicular to the field, there is no change in potential|dw:1361070976684:dw|
yeah that makes sense. cos 90 is zero....

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