## JenniferSmart1 2 years ago Question: I solved the first part and I just need help understanding the second part. This will take me a a few min to write and draw....bear with me...

1. JenniferSmart1

Point charges of $q_1=-5.00 \mu C$$q_2=+3.00 \mu C$$q_3=+5.00\mu C$ located on the x access at $x=-1.00cm$$x=0$$x=+1.00cm$respectively. I calculated $\vec{E}$ on the x axis atx=3.0cm and x=15.0cm and got: $\vec{E}_{p1}=(1.14\times10^8N/C)\hat{i}$ and $\vec{E}_{p1}=(1.74\times10^6N/C)\hat{i}$ Here is where I'm having trouble: ARe there any points on the x-axis where the magnitude of the electric field is zero? If so, where are those points?

2. JenniferSmart1

@Luis_Rivera @phi

3. JenniferSmart1

typo: I meant to write $\vec{E}_{p2}=(1.74\times10^6N/C)\hat{i}$

4. JenniferSmart1

@Luis_Rivera I'll post the answer, can you help me understand it? I have the solution but I'm having a hard time understanding the concept

5. JenniferSmart1

@Jemurray3

6. Jemurray3

|dw:1360443203643:dw|

7. Jemurray3

Think of it this way: if you were just slightly to the left of the particle at x = -1, the electric field would point to the right, correct?

8. JenniferSmart1

correct

9. Jemurray3

Now go a hundred billion miles to the left. The distances between the three would be almost zero, so you'd really only be able to see a single particle of charge -5 + 3 + 5 = 3. Does that argument make sense?

10. JenniferSmart1

Yes, so we would see a particle of charge three. Oh so is $r\rightarrow \infty$?

11. Jemurray3

No. The point of that is if you're close to the one at -1, then the field points to the right. If you're far away, the field points to the left, so there must be some point at which the field is zero.

12. JenniferSmart1

I guess that kinda makes sense....where from here?

13. Jemurray3

The fields from the three particles are: $E_1 = \frac{-5}{(x+1)^2}$ $E_2 = \frac{3}{x^2}$ $E_3 = \frac{5}{(x-1)^2}$ So the equation you should be solving is $E_{total} = E_1 + E_2 + E_3 = 0$

14. JenniferSmart1

$E_1=\frac{q_1}{(x+1)^2}$ this means: the electric field due to a point charge from a positive test particle "x" located at a location (which I'm solving for) $E_{total} = E_1 + E_2 + E_3 = 0$ by solving for x, I've found a place on the x axis where the electric field due to all point charges is zero....

15. JenniferSmart1

makes sense. THanks @Jemurray3 !

16. JenniferSmart1

The positive and negative charges are bothering me...|dw:1360445236763:dw|

17. Jemurray3

What's bothering you?

18. JenniferSmart1

I Would've written this as: $-\frac{q_1}{(x-1)^2}-\frac{q_2}{x^2}-\frac{q_3}{(x+1)^2}=0$

19. JenniferSmart1

because the charge on q_1 is negative

20. Jemurray3

You must also take into account the direction the field is pointing. The field from the q_1 charge is to the right.

21. Jemurray3

Generally speaking $\vec{F} = \frac{kQ}{r^2} \hat{r}$ So Q is negative, but that is compensated by the direction of r hat.

22. JenniferSmart1

That explains the positive sign, what about the x_1 part? why do they have it as x-(-1)?

23. JenniferSmart1

I mean the x-1 part

24. Jemurray3

That is the distance between a point (x) and the point (-1).

25. JenniferSmart1

which should be x-1 though. That doesn't make sense in practice...I think.... Ok let's see (I can't believe I'm struggling with simple arithmetic) $\triangle x=|x_2-x_1|$ so let's say our point is at x_2=-5 Oh I see....nevermind haha yep it's plus (+)

26. Jemurray3

If you ever forget, just say "okay, so where is the distance zero?" and if it matches the location of the actual particle then you're good.

27. Jemurray3

so (x+1) is zero at x=-1 , which is where the particle is.

28. JenniferSmart1

yep, I'll remember that. Thanks again @Jemurray3 !!!