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 one year ago
\[\Delta V=V_bV_a=\frac{\Delta U}{q_0}=\int_{a}^{b}\vec{E}\cdot d\vec{l}\]
 one year ago
\[\Delta V=V_bV_a=\frac{\Delta U}{q_0}=\int_{a}^{b}\vec{E}\cdot d\vec{l}\]

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JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361068864225:dw Is that what it means?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0"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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0"The potential difference \(V_bV_a\) 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)."

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1dw:1361069290883:dw

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1Only 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.

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0Why would E be in the direction from a to b?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1Well, it could be the other way, but usually a is initial and b is final...

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1a positive test charge will move from a to b if the field points from a to b

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0If a and b are the initial and final position of the point charge q....ohhhhh Is the particle moving due to the electric field?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0I see......!!!!dw:1361069892440:dw

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0Sorry @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
 one year ago
Best ResponseYou've already chosen the best response.1yes, 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\[\vec E\cdot\vec r=\\vec E\\\vec r\\cos\thetadw:1361070442394:dw\]

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1oops because of\[\vec E\cdot\vec r=\\vec E\\\vec r\\cos\theta\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0Sorry I didn't quite get that.... which way does the r vector point? tangent to the path?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1dw:1361070847504:dw\(\vec r\) is whatever direction some element of the path \(d\ell\) happens to be pointing in

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1so for all the parts of the path that are perpendicular to the field, there is no change in potentialdw:1361070976684:dw

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0yeah that makes sense. cos 90 is zero....
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