Recent content by Frogger Man

1.194e-11 Got it, again, thanks for your help. Wish I had you as my teacher in school.

1.6x10^-19*1.6x10^-19*8.99x10^9/(9.11x10^-31*(4.6x10^6)^2) That should be better, I used the mass instead of the charge.

I must have a number wrong, when I plug it

9.11x10^-31*1.6x10^-19*8.99x10^9/(9.11x10^-31*(4.6x10^6)^2)

r = q1q2k/(mv^2)

q1q2k/r^2 = mv^2/r

The 1.67*10^6 should be 9.109 × 10-31 On the left i am trying to find the force of attraction, is this the wrong way to go about this problem?

In one model of the hydrogen atom, the electron revolves in a circular orbit around the proton with a speed of 4.6 106 m/s f = mv^2/r F is the force of attraction between the electron and proton m is the mass of the electron v is the velocity it spins at r is the radius...

I got it, 195872.2698. Thanks for your help.

I think you might be right, it should be milli. I will try that.

Fba = k • QA • QB / dBA2 Fba = (8.99*10^9 • (5.85)*10^-6 • (-5.85)*10^-6) / 1.2^2 Fba = (-.30766) / (1.44) Fba = -.21365 Are my numbers off, or just my sign, or both?

They should be in Newtons

Yea, probably should have put that in, Determine the magnitude and direction of the force on each charge.

Two Q = +5.85 mC charges are placed at opposite corners of a square d = 1.2 m on each side, and -Q = -5.85 mC charges are placed at the remaining corners, as shown in the figure. I labeled the square A, B, C, D from top left to bottem left. Fba = k • QA • QB / dBA2 Fca = k • QA • QC /...