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Electric Force Question

  • #1

Homework Statement



Basically, I am given a graph with three charges, q1, q2, and q3 on it and told that the charge on q1 is 22 microcoulombs and the charge on q2 is 19 microcoulombs. I am then told that the force on q1 points in the -x direction which I used to find that the charge on q3 was also 19 microcoulombs since it is the same distance from q1 as q2 is. The coordinates for each charges are as follows:

q1 - (0,1)
q2 - (2,0)
q3 - (2,2)

What I need to find is the magnitude of the force on q1 caused by both q2 and q3.

Homework Equations



I would think just Coulomb's Law.

The Attempt at a Solution



I used the superposition principle and Coulomb's Law and added together the x-components of each charge's individual force and came up with 1.9 Newtons. However, the online answer-checker thing is saying I am incorrect. I also tried doing stuff with finding r(hat) and using that to find the answer, but I still didn't get it right. Any help?
 

Answers and Replies

  • #2
1,137
0
How did you get 1.9N?
 
  • #3
Well, I tried to act as though the x-components of the force from q2 and the force from q3 existed, since the y-component of the force on q1 was 0. This is basically what my math looks like:

((9*10^9)/4)(22*10^-6)(2)(19*10^-6) = 1.88

The 9*10^9 is the proportionality constant and is in N*m^2/C^2. It was factored out of the force addition process.
The 1/4 came from me taking the inverse square of the distance between q1 and the other two charges and was in 1/m^2. It was factored out of the force addition process.
22*10^9 is the charge on q1 and is in C. It was factored out of the force addition process.
The 2 and 19*10^-6 are because of the addition of the charges from q2 and q3 and are also measured in C.

Any idea where I went wrong? =/
 
  • #4
gneill
Mentor
20,793
2,773
This distance between the charges is taken along the line joining them. They're two over and one up (or down). By the way, I suppose we're assuming meters for the distance units? You didn't specify.
 
  • #5
So, I should be using the square root of five as the distance then? And yes, it's in meters, sorry. >.<

Edit: Just remembered, I tried the square root of five as the distance before and got 1.5, but it said that was wrong as well. =/
 
  • #6
1,137
0
Well the distance between q1 and q2 comes out to be sqrt(5).

Now can you tell me the X component of force b/w q1 and q2?
 
  • #7
Umm, would that just end up being .94 N? I basically took ((9*10^9)/(2^2))(22*10^-6)(19*10^-6). Or am I not allowed to use 2 for my distance in that case? =x
 
  • #8
gneill
Mentor
20,793
2,773
So, I should be using the square root of five as the distance then? And yes, it's in meters, sorry. >.<

Edit: Just remembered, I tried the square root of five as the distance before and got 1.5, but it said that was wrong as well. =/
Yup. sqrt(5) is the distance between q1 and the others.

What you are calculating is the force directed along the lines connecting the individual charges. You need to extract the x-component.
 
  • #9
1,137
0
Well its not 0.94!!

You cannot use 2 as distance. Draw figure and tell me what is the smallest distance b/w q1 and q2?
 
  • #10
Extract as in not bother with? So, basically I should use the distance of sqrt(5) between each charge to find the force caused by either q2 or q3 and then multiply it by two since there are two of them and it's symmetrical/same charge? Or am I way off? =x

@Cupid: You mean sqrt(5)? :O
 
  • #11
1,137
0
No dont multiply by 2!!! find the angle between the two forces!!
 
  • #12
1,137
0
or the angle of any1 with X axis, find its X component and then multiply it by 2
 
  • #13
Sorry, I'm really confused about this one. >.< I'm not quite sure where I'm supposed to use the angle, but I think the angle is 26 degrees. How does 6.7 N sound as an answer though? I tried solving for r(hat) and using that and ended up with that answer. =x
 
  • #14
1,137
0
Look, using distance as sqrt(5), you get force b/w q1 and q2 as .7524

and you know that its angle with X axis is 26(26.5 more accurately) ... So its X component will be Force x cos26.5

which is .67!!! (Not 6.7)

So net force will be 2*.67 which is 1.35
 
  • #15
Oh! Thanks, I understand what you're trying to say here now! :D
 

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