- #1

Valentine

- 2

- 0

I am new to the website and I have joined because I just started a Physics class in College and I've never done any sort of Physics and after my first day I was thrown this problem and have been looking at it for a while now and finally gave in for some sort of help.

1. Homework Statement

1. Homework Statement

Particles of charge +70, +48, and -80 μC are placed in a line. The center one is 0.35 m from each of the others. Calculate the net force on each charge due to the other two.

**2. The attempt at a solution**

Before posting, I looked to see if any other posts were similar and I had came across one from 2005 and I've looked at his math:

F = (8.99*10^-9) (48*10^-6) (70*10^-6) / 0.35^2

which comes out to be: 2.46*10^-16

then F = (8.99*10^-9) (70*10^-6) (-80*10^-6) / 0.7^2

which comes out to be: -3.48*10^-16

And the answer I get it:1.4*10^-16

I understand the user, neonerd, had said that he had made a mistake with the -9 on (8.99*10^-9) however I am not making the connection between

"F = (8.99*10^-9) (48*10^-6) (70*10^-6) / 0.35^2

= 2.46*10^-16"

I understand the formula is F = K q1 q2 / r^2

**My professor has k=9x10^9 Nm^2/Coul^2 instead of using k=8.99x10^9 but says that meters and C cancel out so I know why K is different of neonerd's work**

So the formula would start out as

F=(8.99x10^9) q1 q2 / r^2.

R is the distance so 0.35m^2

F=(8.99x10^9) q1 q2 / 0.35m^2

I am not understanding why the q1 and q2 are (70*10^-6) (-80*10^-6) and how neonerd got (2.46x10^2N)

Again I apologize if my misunderstanding is hair-pulling to anyone I just have never seen this and I would appreciate any help.