ELECtric Field Problem PLEASE HELP HELP

In summary, the net electric field is zero at the coordinate (20, 70) due to the opposing electric fields of the two charges.
  • #1
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[SOLVED] ELECtric Field Problem PLEASE HELP ! HELP!

Homework Statement



Two particles are fixed to an x axis: particle 1 of charge q1 = +2.8*10-8 C at x = 20 cm and particle 2 of charge q2 = -4.50q1 at x = 70 cm. At what coordinate on the axis is the net electric field produced by the particles equal to zero?

Do not convert to meters. leave the answers in cm. so we are trying to find the x-coordinate.

i really have no clue where to start.



Relevant equations

E= kQ/ R^2
 
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  • #2
This is similar to the previous problem, and E = kq/r2 for a single charge.

If one is working with cm, then perhaps one is supposed to use cgs/esu units. One can rewrite the charges in esu.

Also, since one is looking for the location of E = 0, the location is not between the charges, since the field points outward from the + charge and inward to the negative charge, so on must find the position where the fields of both charges are equal and opposite.
 
  • #3
Astronuc said:
This is similar to the previous problem, and E = kq/r2 for a single charge.

If one is working with cm, then perhaps one is supposed to use cgs/esu units. One can rewrite the charges in esu.

Also, since one is looking for the location of E = 0, the location is not between the charges, since the field points outward from the + charge and inward to the negative charge, so on must find the position where the fields of both charges are equal and opposite.



Okay i sort of understand what you are saying.
how about this-- i will convert everything to meters but at the end of the problem i will convert it back to cm once i get my answer. that way i won't have to change all my units. okay so zer is closer to Q1. BUT i am not so sure what to do. do i set it like this: E=kq / r^2 = 0 since the net electric field produced by the particles has to equal to zero?
 
  • #4
The net electric field is the sum of the individual electric fields.
 
  • #5
Well if one writes the algebraic equation for the electric field/force, then charges will cancel and one will have an algebraic equation with distance as a variable.

On the x-axis, on which a positive and negative charge are situated, the electric field points from the positive to negative charge between the charges. The field points in the opposite direction on the other side (outside) of either charge.

Keep in mind q2 = -4.50 q1, so q2 is negative, but with a magnitude 4.5 times that of q1. What does that tell one about the strength of the field near each charge, i.e. which field is stronger? What is the implication for the location where E = 0?
 
  • #6
Thank you for answering back, but that made things a lot more confusing. i am sorry, i really just don't understand. could you put it in simpler terms without being as formal. i think that would help me understand the question and how to work the problem.
 
  • #7
the field of Q2 is stronger so 0 is being pulled closer to Q2 more than q1 or basically its closer to q2, but i don't know what i can do with this information.
 
  • #8
Write out the net electric field. As I said before, it's the sum of the two individual fields from the two charges. But, what is the distance r? Imagine, at a distance x from the axis, you have the location you're testing at. So, the distance from the test location to charge q1 is (x - 20) cm. Do the same for the other one.

You will get two solutions, one of which is invalid (which you have to figure out from the charge configuration). What I do is, draw arrows at the three different intervals of where the charge could possibly be. So to the left of q1 and q2, q1's electric field points left, q2's points right. In the middle, they point in the same direction. Can they possibly sum to zero there? Do the same analysis for the right too. Then you will be able to eliminate one of the solutions from the quadratic.
 
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  • #9
Okay that helps somewhat, but i am still very confused.

this is what i got so far:
E=k (2.8*10-8 C) / (x-.20 meters)
E = k (-4.50* 2.8e-8) / (x-.70meters)
 
  • #10
First, Thanks again, you are really helpful!
Second, are those equations right?
Thrid, what should i do after that?
 
  • #11
Keep in mind that the denominator should be squared.

Those are the individual electric fields from the two charges. The net electric field is the sum of them. And this net electric field equals zero. Which is pretty much what I said above.
 
  • #12
Oh right, i forgot about the squaring part, thanks!
ooo okay so 0= [k (2.8*10-8 C) / (x-.20 meters)^2] + [k (-4.50* 2.8e-8) / (x-.70meters)^2]


so i solve for x right?
 
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  • #13
Yeah
 
  • #14
okay hold on, i'll tell you what i got...
 
  • #15
i tried taking the square root of everything but i can't take the square root of a negative number.


okay i know this is simple algebra but how do i solve for x, or in other words isolate x?
 
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  • #16
It's a quadratic. Move everything to one side, and then you should only have three terms, the x^2, x and constant term. Then use the quadratic formula.
 
  • #17
whoa, wait a minute. for some reason I am not getting three terms. this is what i have so far:

0= [251.72 / (x-.2)^2] + [-1132.74 / (x-.7)^2]




so what do you suggest i do after this step?
 
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  • #18
Expand the squared terms.
 
  • #19
okay i did and this is what i got, but i still only have two terms:
0= [251.72 / (x^2-0.4x+.04)] + [-1132.74 / (x^2-1.4x+.49)]
 
  • #20
If it's this:
0= [251.72 / (x-.2)^2] + [-1132.74 / (x-.7)^2]
[251.72 / (x-.2)^2] = [1132.74 / (x-.7)^2].

don't expand (you would be in big mess)!
take square-root of both sides
 
  • #21
hello rootX, thanks let me try that:
[15.865 / (x-.2)] + [33.656 / (x-.7)]
 
  • #22
Your best bet would to be to move each term to separate sides and "cross-multiply". (on second thought, should have done that before expanding.) Then move them to the same side again.

I usually do problems with variables first, then substitute the values in at the very end. Saves a lot of work writing messy decimals down. Then you'd see that the K=9e9 and q_1 term cancels out nicely. (I'd plug in the values before using the quadratic formula). The equation I got at the end was, where c = 4.5, a = 20cm, b = 70 cm:

[tex](1 - c)x^2 + (2 a c - 2 b) x + (b^2 - c a^2) = 0[/tex]

Which is a lot more elegant :p

I don't think taking the square root of both sides is that great an idea. Might lead him to believe there's only one solution to the problem.
 
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  • #23
It is supposed to be a simple problem...
you are making a big mess!

in simple
q1/r1^2 = q2/r2^2

r2/r1 = sqrt(q1/q2)
you need some geometry to figure out r2 and r1

"I don't think taking the square root of both sides is that great an idea. Might lead him to believe there's only one solution to the problem."

OO yea, hez are right. You need to know what's going on if you use my way (I always use it, and it always works).
 
  • #24
i got x by doing rootx's way. of squaring both sides, and then cross-multiplying (like awvvu did)
 
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  • #25
rootX said:
It is supposed to be a simple problem...
you are making a big mess!

in simple
q1/r1^2 = q2/r2^2

r2/r1 = sqrt(q1/q2)

Yeah, my method is just brute force from applying the equations.
 
  • #26
is my answer right?
 
  • #27
physicsbhelp said:
is my answer right?

That's what I get too. But keep in mind how to get the solution if the charges were both positive or both negative. You should also be able to determine the other solution, x = 0.36 if that were the case. Just square rooting both sides and blindly following through will only give you the first solution.
 
  • #28
but rootX if i do r2/r1 = sqrt(q1/q2) i can't because Q2 is a negative number and you can't take the sqrt of a negative number


wait how did you get 0.36?
 
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  • #29
yea, I also got that:
here's how I would do:
q1/r1^2 = q2/r2^2

I know the target can't be in the middle, can neither be on the right side (see the geometry - I learned this in high school: more charge on the right and less charge on left, so less charge wouldn't be able to balance with the more charge)
It must only be on left side. (say x cm from q1)
q1/(x)^2 = 4.5q/(x+50)^2 -- simply ignoring the signs (I don't care about them)

(x+50)/x = sqrt(4.5)

x = 44.5 cm from q1

use geometry, and you will find it's -24.5 cm.
 
  • #30
physicsbhelp said:
wait how did you get 0.36?

There's always two solutions to a quadratic. If you did it my way and expanded, the quadratic equation would give you -0.246 and 0.360. The 0.360 answer is invalid in this case because the charges had opposite signs, but is valid if they were both the same sign.

You need to eliminate one of the solutions using the method I said earlier. rootx's way works fine too if you don't blindly apply it.
 
  • #31
well wait, it would be a negative right? and that is the right x-coordinate right?
 
  • #32
Yes, the two solutions to the quadratic are -24.6 and 36.0 cm, not +- 24.6 and 36.0cm.
 
  • #33
THANK YOU thank you THANK YOU thank you THANK YOU VERRRYYYYY MUCH!
 
  • #34
YOU TWO PEOPLE ARE THE BEST OF THE Best! GOD BLESS
 

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