- #1
Valentine
- 2
- 0
Hello all,
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
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:
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.
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
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.