Coulumb 's Law [Archive] - Physics Forums

ACLerok

Sep3-04, 02:48 PM

jamesrc

Sep3-04, 02:53 PM

I didn't check your numbers, but I can tell the sign of the charge you gave can't be right. The first charge is attracting the second charge, so they must have opposite signs.

nrqed

Sep3-04, 03:04 PM

a charge of -0.55 * 10^-6 C exerts an upward force of magnitude .250 N on an unknown charge that is .3 meters directly below it. i am to find the unknown charge but I keep getting -5 *10^-6 which I keep getting told its wrong. Can someone give me a hand here? thANKS

As someone else mentioned, your sign is wrong. Other than that, your value is close but it looks like you rounded off too much. Given the data, you should have kept two sig figs.

Pat

ACLerok

Sep5-04, 05:25 PM

As someone else mentioned, your sign is wrong. Other than that, your value is close but it looks like you rounded off too much. Given the data, you should have kept two sig figs.

Pat

alright thanks guys.. ill check it again.

ACLerok

Sep6-04, 09:24 PM

As someone else mentioned, your sign is wrong. Other than that, your value is close but it looks like you rounded off too much. Given the data, you should have kept two sig figs.

Pat

well there ya go. the answer was 4.55*10^(-6) C. THANKS!

oh and BTW.

A particle of charge 4.96 nC is placed at the origin of an xy-coordinate system, and a second particle of charge 2.02 nC is placed on the positive x-axis at 3.96cm . A third particle, of charge 6.01cm is now placed at the point 3.96cm , 3.01cm. I found the x component of the total force exerted on the third force by the first two. But now i must find the y component of the total force.

The components can be negative or positive correct indicating direction or should i disregard that when adding the forces together?
I'm having alot of trouble finding the y component of the Force exerted by charge 2 on 3. Can someone help me out?

The answers I did find are
F(1on3)x=8.685*10^-5
F(2on3)x=0
F(1on3)y=-6.601*10^-5
F(2on3)y=9.641*10^-5

THANKS!!!!