# Electrostatic force between more than 2 charges

1. Jan 20, 2008

### moonlight13

hi
i dont know if this has been posted or not but i am sort fo stuck on this one

What is the force F_vec on the 1 nC charge at the bottom? View Figure
Write your answer as two vector components, separated by a comma. Express each component numerically, in newtons, to three significant figures.

the figure looks like this

attempt to this problem:

i found out the forces individually and added them together .. but it didnt work

x-component:
6nC = 0
2nC = 5.09 x 10$$^{-5}$$ each

y-component:
6nC = 2.16 x 10$$^{-5}$$
2nc = 5.09 x 10$$^{-5}$$ each

when i add them: 1.02 x 10 $$^{-5}$$ $$\hat{x}$$ , 7.76 x 10$$^{-5}$$ $$\hat{y}$$

but that was wrong.. i also tried the x-comp to be 0 .. but that was wrong too

thanx in advance for the help

2. Jan 20, 2008

### Staff: Mentor

How does the x-component of the force exerted by one 2 nC charge compare to that of the other?

3. Jan 20, 2008

### moonlight13

they are opposite of each other .. which means there is no net x- component .. but thats not the right answer tho..

4. Jan 20, 2008

### TVP45

I see 2 things you might do differently to help.
Make sure you're doing vectors (right is +, up is +). Keep the vector notation throughout.
Check your use of scientific notation. If you're unsure, have your calculator switch it to regular decimal.

5. Jan 20, 2008

### moonlight13

i tried a similar question, with 1 + and 1 - 2nC charge .. and that worked out
but somehow i cant get this working ..

i am not too sure if the y - component it right as well or not .. even tho it seems right

6. Jan 20, 2008

### Staff: Mentor

Show how you calculated the vertical components of the forces from the 2 nC charges.

7. Jan 20, 2008

### Staff: Mentor

What are the directions (signs) of these components?

8. Jan 20, 2008

### moonlight13

y-comp = Esin$$\vartheta$$
r= 5x10^-2
E = Kq$$_{1}$$q$$_{2}$$ / r$$^{2}$$
= (9.0x10^9)(2x10^-9)(1x10^-9)/ ( 2.5x10^-3)
E= 7.2x10^-6

y-comp = E sin(45) = 5.09x10^-6

9. Jan 20, 2008

### moonlight13

6nC = +y, no x
2nC = +y, +x (on the right)
2nC = +y, -x (on the left)

and thats why my understand was that the x components would cancel out .. but i am not sure whats going on

10. Jan 20, 2008

### Staff: Mentor

Realize that the 2 nC charges are positive and the 6 nC charge is negative. Which way do their forces on the + 1 nC charge act?

11. Jan 20, 2008

### moonlight13

so the 2 +y .. should be -y ..?

12. Jan 20, 2008

### moonlight13

i did the calc again and i get

x = 0
y = 1.14x10^-5 ..

doe sthat seem right?

13. Jan 20, 2008

### Staff: Mentor

Correct. Since the charges repel, the y component of force is downward and thus negative.

14. Jan 20, 2008

### moonlight13

thank you ., thank you ... thank you..

15. Jan 20, 2008

### Staff: Mentor

What's the distance between the charges?

16. Jan 20, 2008

### moonlight13

what you said abt the charges repelling .. this answer is right ..

thank you so much once again

17. Jan 20, 2008

### Staff: Mentor

You are most welcome.

Never mind. You had written the distance squared.