Electric Force, Equilibrium Config of Charges

[arl146]

Homework Statement

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.

Homework Equations

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

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 Cmy 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 isn't it just (2*k*q^2)/r^2

 

[Infinitum]

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

 

[arl146]

i actually don't 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 .. isn't 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 ..

 

[arl146]

it doesn't 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

 

[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.

 

[arl146]

adding those 2 vectors equals the 'hypotenuse' .. i don't know what you want for an answer

 

[Infinitum]

adding those 2 vectors equals the 'hypotenuse' .. i don't know what you want for an answer

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

 

[arl146]

sqrt(2)*F

 

[Infinitum]

sqrt(2)*F

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...

 

[arl146]

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

 

[Infinitum]

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

How did you get the second term from!?

When they're perpendicular,

F + F = sqrt(2) F, that's it.

 

[arl146]

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

 

[Infinitum]

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

Well, there you go

 

[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) ..

 

[Infinitum]

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

Are you really sure...?

 

[arl146]

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

ohh yea duhh i don't know what i was thinking for that part .. i think i got the radical on the bottom mixed up

 

[arl146]

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

 

[Infinitum]

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

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

 

[arl146]

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