# Homework Help: Electric Force, Equilibrium Config of Charges

1. May 14, 2012

### arl146

1. The problem statement, all variables and given/known data
Find the charge Q that should be placed at the centre of the square of side 8.50E+0 cm, at the corners of which four identical charges +q = 3 C are placed so that the whole system is in equilibrium.

2. Relevant equations
F=(k*q1*q2) / r^2

3. The attempt at a solution
I know the solution is something like this:

Let r = side of square. That's 8.5 cm, right?
Without the center charge, F1 (on each charge, away from center) = kq^2/(sqrt(2)*r)^2 (diagonal charge) + 2*sqrt(0.5)*kq^2/r^2 (2 adjacent charges) =
k(q/r)^2*(0.5+sqrt(2)) = 2.1430705E13 N
F1 = -F (center charge Q to each corner charge q) = kqQ/(r*sqrt(0.5))^2 = 2kqQ/r^2
==> Q = -(F1)r^2/(2kq) = -(0.5+sqrt(2))q/2 = -2.87132 C

my only question is this:
why does the F of the 2 adjacent charges = 2*sqrt(0.5)*kq^2/r^2
where does the 2*sqrt(1/2) come from? why isnt it just (2*k*q^2)/r^2

2. May 15, 2012

### Infinitum

Forces are vectors. You can't add them like normal scalars. Draw a diagram and split components, and you'll see why

3. May 16, 2012

### arl146

i actually dont get this whole thing at all lol ...

so when you break it down, F1 = F12 + F13 + F14
F12 and F14 are the adjacent charges. aren't those just kq^2/r^2 for each?
and the F13 is the diagonal charge .. isnt that just kq^2/r^2 but in the x dir it has a cos(theta) attached to it and for the y dir it has a sin(theta) attached ..

4. May 16, 2012

### arl146

it doesnt make sense at all. F12 should just be (k*q^2)/r^2 and same with F14 .. it doesn't make sense to me why F12+F14 = (2*sqrt(0.5)*k*q^2)/r^2

5. May 16, 2012

### Infinitum

Do you know addition of vectors? Can you tell me what the sum of the two vectors in the attachment would be? Assume the magnitude of both the forces is F, and they are perpendicular.

#### Attached Files:

• ###### force.bmp
File size:
169.4 KB
Views:
130
6. May 16, 2012

### arl146

adding those 2 vectors equals the 'hypotenuse' .. i dont know what you want for an answer

7. May 16, 2012

### Infinitum

Yes, its the hypotenuse. What is the magnitude of that resultant sum?

8. May 16, 2012

### arl146

sqrt(2)*F

9. May 16, 2012

### Infinitum

Yep. Now what you have in your original question is,

F12 = F14 = (kq2)/r2 = F

And they are perpendicular to each other....Apply the same concept you used for my question...

10. May 16, 2012

### arl146

then F12 + F14 = sqrt(2)*(kq^2)/r^2 + sqrt(2)*(kq^2)/r^2 = 2*sqrt(2) (kq^2)/r^2 ...

11. May 16, 2012

### Infinitum

How did you get the second term from!?

When they're perpendicular,

F + F = sqrt(2) F, thats it.

12. May 16, 2012

### arl146

oh yea right. so just sqrt(2)*(kq^2)/r^2 so whats that mean now

13. May 16, 2012

### Infinitum

Well, there you go

14. May 17, 2012

### arl146

well you didnt answer my question ... in the problem's answer i posted in my original first post, which gave me the right answer, it says that F12+F14 = 2*sqrt(0.5)*kq^2/r^2

2*sqrt(1/2) does not equal sqrt(2) ..

15. May 17, 2012

### Infinitum

Are you really sure....?

16. May 17, 2012

### arl146

maybe they made a typo since it still works out *shrugs* thats all the reason why i was confused ... thanks though :)

ohh yea duhh i dont know what i was thinking for that part .. i think i got the radical on the bottom mixed up

17. May 17, 2012

### arl146

ohhh yea duh idk what i was thinking for that. mixed up the radical on the bottom

18. May 17, 2012

### Infinitum

$2\frac{1}{\sqrt{2}} = \sqrt{2}$

19. May 17, 2012

### arl146

haha yea like i said mixed up the radical. wasnt thinking to take out the radical on the bottom lol