## anonymous 4 years ago Three 1.0 nC charges are placed as shown in the figure. Each of these charges creates an electric field vector E at a point 3.0 cm in front of the middle charge.What are the three fields vector E1, vector E2, and vector E3 created by the three charges? Write your answer for each as a vector in component form.

1. anonymous

I mainly need help on starting it.

2. TuringTest

Got a picture?

3. anonymous

sorry yes there is just hold on

4. anonymous

|dw:1327883171291:dw|

5. anonymous

q1,q2q3 all equal 1nC

6. TuringTest

so you need the field vector from each charge at that point on the right?

7. anonymous

Yes

8. TuringTest

|dw:1327883385609:dw|The easiest one to find is the field due the the second charge$\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\overrightarrow{r}=<0,\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}>$You can just plug the numbers in, and you know the angle will be 0 because the vector points in the +x direction. Got it so far?

9. anonymous

I do understand that so far but it says to put it in vector component form. How would i do that? Like (Ex, Ey). I know Ey should be zero but how would you find Ex?

10. TuringTest

this is component form:$<0,\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}>$^^^ ^^^ Ex=0 Ey=that^

11. anonymous

ohhh ok

12. TuringTest

sorry Ex and Ey are backwards above...

13. TuringTest

$<\frac{1}{4\pi\epsilon_0}\frac{q}{r^2},0>$ Ex^^ ^^Ey=0 I just noted that it says that the point is directly to the right of q2, so since the sine of 0=0 the y-component disappears.

14. anonymous

now how do you find the components q1?

15. TuringTest

let's put a grid on our point so we can do a little trigonometry:|dw:1327884328041:dw|The next one I want to find is the vector due to q3. You will see why soon.

16. anonymous

ok

17. TuringTest

notice all the geometry we can do here, and all the places this angle theta turns up|dw:1327884443364:dw|we can find r with the pythagorean theorem, and the angle with the inverse tangent.

18. TuringTest

$r=\sqrt{3^2+1}$$\theta=\tan^{-1}(\frac13)$so we have$\overrightarrow{E_3}=\frac{1}{4\pi\epsilon_0}\frac{q_3}{r^2}\overrightarrow{r}=<\frac{1}{4\pi\epsilon_0}\frac{q_3}{r^2}\cos\theta,\frac{1}{4\pi\epsilon_0}\frac{q_3}{r^2}\sin\theta>$so this can be solved quickly as well now

19. TuringTest

^I suppose you have to put the centimeters in terms of meters above actually

20. TuringTest

$r=\sqrt{0.03^2+0.01^2}$$\theta=\tan^{-1}(\frac13)$

21. anonymous

Ok, i understand all of that so far.

22. TuringTest

well if you have the radius and angle you can solve that guy, right? now E1 will be simple, because by symmetry it will have the same vector as E3, only the y-component will be negative$\overrightarrow{E_1}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\overrightarrow{r}=<\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\cos\theta,-\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\sin\theta>$where theta is the angle we used earlier. Same r as well.

23. TuringTest

|dw:1327885322988:dw|

24. anonymous

Thanks for helping this the set up. I'll let you know if it doesn't work on the homework. (it can be finicky sometimes)

25. TuringTest

26. anonymous

For some reason it keeps saying I'm wrong for my q2 answer. I wrote that the answer was (1,0). I'm pretty sure it's right but just want to check with you (the answer is in kiloNewtons by the way).

27. TuringTest

let me bust out the calculator...

28. anonymous

thanks for doing that :).

29. TuringTest

wait, the answer is in V/m or N/c or something, not just N that's force

30. anonymous

kN/c

31. TuringTest

32. anonymous

Yep that worked. thanks, i now see what i did wrong.

33. TuringTest

Have you tried the others yet? if not be careful of the units is all I can say...

34. anonymous

yes, they all worked. Thank you very much

35. TuringTest