Find the Distance for Net Zero Force: Two Charges at Origin and x=L

cmcdonald29
Messages
4
Reaction score
0
1.)Two charges are fixed in location: charge q1= +8e is located at the origin and charge q2= -2e is located on the x-axis at x=L. At what point can a proton be placed so that it has net zero force acting on it?


2.)F=kq1q2/r^2


3.)I tried to do F=0
K = 8.988 X 10^9 (constant)
q1 = (8 protons) X (1.602 X 10^-19) ---> (individual charge of one proton)
q2 = (2 electrons) X (1.602 X 10^-19)

I used the equation above then to find r^2; however, as you can see F = 0 X r^2 would give me zero and you cannot divide by 0 to find r. I am confused on how to find the distance.
 
on Phys.org
Hi cmcdonald, welcome to PF!

cmcdonald29 said:
1.)Two charges are fixed in location: charge q1= +8e is located at the origin and charge q2= -2e is located on the x-axis at x=L. At what point can a proton be placed so that it has net zero force acting on it?


2.)F=kq1q2/r^2


3.)I tried to do F=0
K = 8.988 X 10^9 (constant)
q1 = (8 protons) X (1.602 X 10^-19) ---> (individual charge of one proton)
q2 = (2 electrons) X (1.602 X 10^-19)

I used the equation above then to find r^2; however, as you can see F = 0 X r^2 would give me zero and you cannot divide by 0 to find r. I am confused on how to find the distance.
Don't do plug-and-play with the formulas, think about the problem first.

I suggest drawing a picture to illustrate the situation, and think about the forces that are acting on the proton. Also, can you figure out without any calculation in what region you expect the proton to be: ##x \leq 0##, ##0 < x \leq L##, or ##x > L##?
 
I would expect the charge to be located on the opposite side of the electron the exact same distance from the origin proton location.
 
cmcdonald29 said:
I would expect the charge to be located on the opposite side of the electron
That's correct.

cmcdonald29 said:
the exact same distance from the origin proton location.
We'll see about that one after you've solved the equations...

So, put the proton in your drawing and write the formula for all the forces acting on it.
 
I'm still a bit confused on how to actually approach the problem itself. I don't understand what equations I should be using to get it started?
 
The equation you gave in point 2 of the OP.
 
Yes, but I am confused on how I would plug into the equation as putting a 0 force in with not let me work out any distance.
 
cmcdonald29 said:
F=kq1q2/r^2
That equation is for two charges. What happens when you have three?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
9
Views
3K
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
9
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K