# Electrostatic force

1. May 29, 2008

### catie1981

[SOLVED] electrostatic force

1. The problem statement, all variables and given/known data
The figure shows an arrangement of four charged particles, with angle θ = 33.0 ˚ and distance d = 3.00 cm. Particle 2 has charge q2 = 8.00 × 10-19 C; particles 3 and 4 have charges q3 = q4 = -1.60 × 10-19 C. What is the distance D between the origin and particle 2 if the net electrostatic force on particle 1 due to the other particles is zero?

2. Relevant equations

F= k [(q1*q[2,3,4])/ r^2] (*cos theta, if applicable)

3. The attempt at a solution
I know that the forces on particle 1 from 2, 3, and 4 are going to cancel....and so I've reached this point

2[9e9(q1*-1.6e-19)/((.03/cos33)^2)cos33]=9e9(q1*8e-19)/((.03+D)^2)

I guess my issue is there are two unknowns in the equation, so I'm stuck. After searching the web, and using the publishers tutorial, I am still utterly lost. The tutorial did ask this very confusing question as well, which I had no idea how to determine the answer for that, soo help would be appreciated for that too. The question was "What multiple of q1 gives the magnitude of the force on particle 1 due to particle (2,3,4)?" The answer was
.000001122312 N/C. WHY??! Any help at this point will be greatly appreciated! Thanks,
Catie
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### q20.jpg
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2. May 31, 2008

### catie1981

Is my question impossible to answer? I hope not!! Maybe informing any would-be helpers that the figure mentioned in the problem is the attached figure might help? I still can't get any further with it, so don't give up on me yet!! Thanks

3. May 31, 2008

### konthelion

Hello.
By Coloumb's law, $$F=\frac{kq_{1}q_{2}}{r^2}$$

Since you are given that $$q_{3}=q_{4}$$,

$$F=\frac{kq_{3}q_{4}}{r^2}$$
$$\Rightarrow F= \frac{q_{3}^2}{r^2}$$

Plug in for $$q_{3}$$ and $$r^2$$.

Next, after you have found F you need to find the coordinates.
We know that
$$F_{x} = \frac{kq_{3}q_{4}cos{\theta}}{r^2}$$
$$F_{y} = \frac{kq_{3}q_{4}sin{\theta}}{r^2}$$

After you've found the coordinates, use the distance formula to find the distance from $$q_1 \rightarrow q_2$$

Correct me if I'm wrong.

Last edited: May 31, 2008
4. May 31, 2008

### Redbelly98

Staff Emeritus
Catie,

In your equation (which seems correct to me), look for common factors on both sides which can be divided out. That will simplify things.

5. Jun 1, 2008

### catie1981

konthelion, not sure how to use the distance formula to obtain that value using information on q3 and q4, but thank you for the help.
Redbelly, I can see that potentially the q1 variable and 9e9 could cancel, but after my lousy attempts at algebraically solving for D, I found a negative number to take the square root of, and, ha ha ha, you can't do that! So, I am stuck again, although, your tip was especially helpful! Thanks!

6. Jun 1, 2008

### Redbelly98

Staff Emeritus
I'll rephrase my earlier remark, and say the equation looks almost correct.

Hint: since you are being appropriately careful with your plus and minus signs on the forces, remember that the forces from the different charges should sum to zero. How does that change the equation you wrote?

7. Jun 1, 2008

### dynamicsolo

You didn't get an answer for some time because you made your diagram (without which your question cannot be worked on) an attachment to your post. The anti-spam strategy on this Forum is that a moderator has to examine any such attachments before they will be permitted to display here. (And it looks like it was more than 24 hours before it got passed -- since most schools are out right now, the pace of activity here is going to be relatively slow...)

In the future, you will get a quicker response if you use a free hosting service to post your picture to the 'Net and then use the link in your post on the Forum. That allows immediate access for the helpers here and does not require a judgement by the moderators...

8. Jun 1, 2008

### catie1981

dynamicsolo, thanks for the advice. I was beginning to think that no one wanted to help me :( but I knew better!
RedBelly, ok, so should I set up the equation so that all the variables are on one side and set it equal to zero?? Subtract force from particle two from the forces from particles three and four?? Arrgghh math is my adversary!!

9. Jun 1, 2008

### Redbelly98

Staff Emeritus
I would put it this way: Add together the forces from particles two, three, and four, and set it equal to zero.

Note: there will be minus signs appearing in the force sum, for any of the negative charges.

10. Jun 1, 2008

### catie1981

OK....but what about the two variables present in the equation? I don't have a value for q1, so I can't figure the forces on that particle....I'm sorry that I'm such a dunce!

11. Jun 1, 2008

### Redbelly98

Staff Emeritus
You might get lucky again with the new equation ...

12. Jun 1, 2008

### catie1981

would my new equation happen to look like this?:

2[9e9(q1*-1.6e-19)/((.03/cos33)^2)cos33]-[9e9(q1*8e-19)/((.03+D)^2)]=0

or am I on the totally wrong track?

13. Jun 1, 2008

### Redbelly98

Staff Emeritus
You're close! But we should add the forces instead, since they sum to zero:

2[9e9(q1*(-1.6e-19))/((.03/cos33)^2)cos33] + [9e9(q1*8e-19)/((.03+D)^2)]=0

As I said a while back, look for common factors which can be "divided out".
Also, remember the first expression will be negative owing to the -1.6e-19 value.

14. Jun 1, 2008

### catie1981

when you say "divided out" you mean factored out, right? and the resulting equation will have only the D variable left for something like this:

[-1.6e-19/((.03/cos33)^2)]cos33 + 8e-19/(.03+D)^2 = 0

(my math could be really off, as I am a "B" math student so....)
(I have to log off now, as my shift is over at work, and I have to go home....)Thank you so much for your help!!!

15. Jun 2, 2008

### Redbelly98

Staff Emeritus
Yes, I meant factored out. The equation looks good, you could also factor out the (e-19).

16. Jun 3, 2008

### catie1981

woohoo!!! I got this one!!! YAY!!! thanks so much for directing in the correct path!

17. Jun 3, 2008

### Redbelly98

Staff Emeritus
You're welcome.