Net Force on Center Charge in Y-Direction in Square of Charges

[learningphysics]

but y the q^2, shouldn't it be q_{1}q_{2}?

edit: oh wait i see what you did. shouldn't r=.332?

I'm getting r^2 = (0.235/2)^2 + (0.235/2)^2

 

[Winzer]

mmm...so what is wrong with:
F= \frac{qKCos(\theta)}{r^2}[-1.4e^-6-1.75e^-6+7e^-7+2.8e^-6]
i chose the negitives because -x direction.

edit: for the x direction

 

[learningphysics]

mmm...so what is wrong with:
F= \frac{qKCos(\theta)}{r^2}[-1.4e^-6-1.75e^-6+7e^-7+2.8e^-6]
i chose the negitives because -x direction.

edit: for the x direction

Can you show how you get your numbers? I'm not getting those numbers...

 

[Winzer]

well A=2*3.50e-7=7.0e-7
B=4*3.50e-7=1.4e-6...

 

[learningphysics]

well A=2*3.50e-7=7.0e-7
B=4*3.50e-7=1.4e-6...

Oh, I see now... Ok... but you should be taking the y-components, not the x-components... so the 1.4e-6 and 7e-7 should have the same sign. And the other two should have the same sign.

 

[Winzer]

ok so for y this should work right?
F= \frac{qKSin(\theta)}{r^2}[-1.4E^-6-1.75E^-6+7E^-7+2.8E^-6]

 

[learningphysics]

ok so for y this should work right?
F= \frac{qKSin(\theta)}{r^2}[-1.4E^-6-1.75E^-6+7E^-7+2.8E^-6]

No, that won't work.

 

[Winzer]

but y is what i am wondering, beside the wrong value.

 

[learningphysics]

but y is what i am wondering, beside the wrong value.

Switching from cos to sin like that will only work if you use a different angle for each charge (measure the angle from the +x - axis)...

 

[Winzer]

ok,ok. But how do I correct the equation?!

 

[learningphysics]

ok,ok. But how do I correct the equation?!

The quantity in your square brackets should be:

[-1.4E^-6+1.75E^-6-7E^-7+2.8E^-6]

I just fixed the signs... sin(45) = cos(45) so it doesn't matter if you use sin or cos.

 

[Winzer]

did u get .197N? i didn't

 

[learningphysics]

The sum of the square brackets is: 2.45E-6

so we want:
kq/r^2* sin(45) (2.45E-6)

r^2 = (0.235/2)^2 + (0.235/2)^2 = 0.0276125

so we get:

[9E9(3.5E-7)/(0.0276125)]*sin(45)*(2.45E-6) = 0.1976N

 

[Winzer]

lol..ok now i get. I must have misinterperted the diagram to have small box lengths of.235...