## elica85 3 years ago the potential as a function of position x is shown in the graph...

1. elica85

|dw:1332350001701:dw|

2. elica85

i tried to draw a smooth curve

3. elica85

is the electric field zero at x=0? is the max magnitude at 15 or 5cm?

4. ash2326

Electric Field is defined as $E=-\frac{dv}{dx}$ v= potential so E is the negative of the slope of the potential curve. Do note that E found from this relation is the E in x - direction. At x=0 the curve is parallel to x -axis, that is it's constant so E=0 At x=5 the slope is negative and magnitude wise the highest. So E will be max at x=5 cm

5. elica85

how do you determine that the curve is parallel to x axis @ x=0? it doesn't look like it on my graph but in the actual graph, it looks parallel at x=10 too

6. elica85

i assumed E was zero at x=0 since d(v=0) should be zero so E=(d0)/dx=0

7. ash2326

No if v=0 doesn't mean that E=0 but the negative of slope of V is E, Yeah at x=10 also the slope is zero. So there also E is zero

8. elica85

hmm still confused. i need the negative of the slope of V to find where E is zero? or to find the max magnitude? and which direction is the E at the max point?

9. elica85

|dw:1332352015313:dw| is this what the graph of E look like?

10. ash2326

Ok Elica, Yeah Electric field is the negative of the voltage's slope. When we have Voltage constant or parallel to axis. Then it's slope is zero and consequently E is zero

11. elica85

ok i understand. so is the graph i did for E right?

12. ash2326

If V is cos then E will be sine

13. ash2326

x=0, v=constant so E will be zero

14. elica85

ok so it's not right. how do i determine the direction of E at x=0 or x=10?

15. ash2326

|dw:1332352349621:dw|

16. ash2326

E is in the x direction always:)

17. elica85

so right either way

18. ash2326

Yeah, when E is negative that means it's in -x direction

19. elica85

thank you so much

20. ash2326

Welcome, Did you understand?

21. elica85

yes, that was great help