# Point charges on square - find magnitude of force

1. Feb 3, 2004

### pringless

Four point charges, each of magnitude 12.92 x 10^-6 C, are placed at the corners of a square 65.5 cm on a side. Given K_e = 8.98755 x 10^9 Nm^2/C^2. If three of the charges are positive and one is negative, find the magnitude of the force experienced by the negative charge. Answer in units of N.

can someone help me out with this? thanks

2. Feb 3, 2004

### Staff: Mentor

Can you find the force that each positive charge exerts on the negative charge? (Hint: Coulomb's law) Draw yourself a picture to keep the direction of each force straight. Now add the force vectors up. (Another hint: take advantage of fact that the charges are symmetrically distributed; you should be able to tell which direction the net force points.)

3. Feb 3, 2004

### pringless

im not sure what u mean
this is what i tried

i found the magnitude of each vector using kqq/r^2
then i squared all 3 magnitudes and added, and then took the square root. but its wrong.
what can i do?

4. Feb 3, 2004

### Staff: Mentor

You found each separate force vector. Excellent. Now describe the direction of each vector. You must first add these vectors before you can find the magnitude of the resultant force. I have attached a picture of the four charges that may prove helpful. Why don't you use the picture to describe the forces.

5. Feb 3, 2004

### Staff: Mentor

Oops... here it is.

#### Attached Files:

• ###### charges.doc
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Last edited: Feb 3, 2004
6. Feb 3, 2004

### pringless

how do i recieve the attachment?

7. Feb 3, 2004

### Staff: Mentor

Oh well, the attachment doesn't seem to be working. Why not draw one of your own?

8. Feb 3, 2004

### pringless

i think i have the directions correct, how do i add the vectors here?

9. Feb 3, 2004

### Staff: Mentor

To add the vectors, I suggest finding their components. Add the components, then you can find the magnitude of the resultant.

10. Feb 4, 2004

### pringless

i keep getting it incorrect
i would use cos45 for the x compoenent and sin45 for the y right?
then i add those components to the other two magnitudes?

11. Feb 4, 2004

### Staff: Mentor

Not quite. Let's do it step by step. It looks like my diagram did finally make it into the system; go to my previous posts and take a look.

Charges at A, B, and C exert forces on the negative charge at D. Tell me the x and y components of each of those forces.

12. Feb 4, 2004

### pringless

F_db*cos45 and F_db*sin45 ?

13. Feb 4, 2004

### Staff: Mentor

Which force is this? Do it systematically for each force:

What force does A exert on D: magnitude and x & y components.
What force does B exert on D: magnitude and x & y components.
What force does C exert on D: magnitude and x & y components.

14. Feb 4, 2004

### pringless

this is my view of it.

A+ B+

C- D+

F_dc = kqq/R^2
F_bc = kqq/(sqr(x))^2 [x = distance of side of square]
F_bc_x = kqq/(sqr(x))^2 * cos45
F_bc_y = kqq/(sqr(x))^2 * sin45

15. Feb 5, 2004

### Staff: Mentor

I just realized that you were drawing your own diagram. D'oh!
Here's what I get (using your diagram) for the forces on C:
$$F_{AC} = \frac{kq^2}{x^2}$$ (magnitude)
$$F_{BC} = \frac{kq^2}{2x^2}$$ (magnitude)
$$F_{DC} = \frac{kq^2}{x^2}$$ (magnitude)

Now the components:
$$F_{AC-x} = 0; F_{AC-y} = \frac{kq^2}{x^2}$$
$$F_{BC-x} = \frac{kq^2}{2x^2}cos(45); F_{BC-y} = \frac{kq^2}{2x^2}cos(45)$$
$$F_{DC-x} = \frac{kq^2}{x^2}; F_{DC-y} = 0$$

$$F_{x} = \frac{kq^2}{x^2}(1 + \frac{cos(45)}{2})$$
$$F_{y} = \frac{kq^2}{x^2}(1 + \frac{cos(45)}{2})$$

Now the magnitude of the resultant:
$$F_{net} = \sqrt{2}\frac{kq^2}{x^2}(1 + \frac{cos(45)}{2})$$

Plug in the numbers and you'll get your answer. (Unless I goofed.)

16. Feb 5, 2004

### pringless

for the part where you showed the y component do you mean sin45 or is it really cos45?

17. Feb 5, 2004

### Staff: Mentor

sin(45)=cos(45), so it doesn't matter.

18. Feb 5, 2004

### pringless

hah oh right..forgot

thanks for ur patience and help doc, ill reply back to tell you if the answer is correct

19. Feb 5, 2004

### pringless

It's correct. Thanks a bunch Doc.

20. Feb 5, 2004

### Staff: Mentor

You are welcome.