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JenniferSmart1
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\[\Delta V=V_bV_a=\frac{\Delta U}{q_0}=\int_{a}^{b}\vec{E}\cdot d\vec{l}\]
 one year ago
 one year ago
JenniferSmart1 Group Title
\[\Delta V=V_bV_a=\frac{\Delta U}{q_0}=\int_{a}^{b}\vec{E}\cdot d\vec{l}\]
 one year ago
 one year ago

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

JenniferSmart1 Group TitleBest 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
 one year ago

JenniferSmart1 Group TitleBest 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)."
 one year ago

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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

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

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

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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
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

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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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\[\vec E\cdot\vec r=\\vec E\\\vec r\\cos\thetadw:1361070442394:dw\]
 one year ago

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

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

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

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

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