Here's the question you clicked on:

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

• This Question is Closed
1. JenniferSmart1 Group Title

|dw:1361068864225:dw| Is that what it means?

2. JenniferSmart1 Group Title

"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

3. JenniferSmart1 Group Title

"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)."

4. TuringTest Group Title

|dw:1361069290883:dw|

5. TuringTest Group Title

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.

6. JenniferSmart1 Group Title

Why would E be in the direction from a to b?

7. TuringTest Group Title

Well, it could be the other way, but usually a is initial and b is final...

8. TuringTest Group Title

a positive test charge will move from a to b if the field points from a to b

9. JenniferSmart1 Group Title

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?

10. JenniferSmart1 Group Title

I see......!!!!|dw:1361069892440:dw|

11. JenniferSmart1 Group Title

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?

12. TuringTest Group Title

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$\vec E\cdot\vec r=\|\vec E\|\|\vec r\|\cos\theta|dw:1361070442394:dw|$

13. TuringTest Group Title

oops because of$\vec E\cdot\vec r=\|\vec E\|\|\vec r\|\cos\theta$

14. JenniferSmart1 Group Title

Sorry I didn't quite get that.... which way does the r vector point? tangent to the path?

15. TuringTest Group Title

|dw:1361070847504:dw|$$\vec r$$ is whatever direction some element of the path $$d\ell$$ happens to be pointing in

16. TuringTest Group Title

so for all the parts of the path that are perpendicular to the field, there is no change in potential|dw:1361070976684:dw|

17. JenniferSmart1 Group Title

yeah that makes sense. cos 90 is zero....