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

[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...