## JenniferSmart1 2 years ago $\Delta V=V_b-V_a=\frac{\Delta U}{q_0}=-\int_{a}^{b}\vec{E}\cdot d\vec{l}$

1. JenniferSmart1

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

2. JenniferSmart1

"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

"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

|dw:1361069290883:dw|

5. 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.

6. JenniferSmart1

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

7. TuringTest

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

8. TuringTest

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

9. JenniferSmart1

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

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

11. JenniferSmart1

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

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

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

14. JenniferSmart1

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

15. TuringTest

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

16. TuringTest

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

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