# Homework Help: Need some help with forces of changed particles (coulomb's Law)

1. Jan 26, 2005

### neonerd

I need somebody to explain to me specifically "Forces of changed particles & Coulomb's Law", which was covered in a class I missed yesterday.

Basically, I know that 8.99 x 10^-9 is a constant used in the beginning of the equation, and then after you put (q1 and q2), which i have no clue what it means, but i know it's the numbers that correspond to the particles , and divide everything by distance ^2

I saw some people use 1.6 x 10 ^-19 for some problems, and the teacher said it was a constant too...I just have no clue when to use this constant, and when not to use it. It was used in some problems, but not in others

Here's an example of one of the problems we have to do:

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 force on each charge due to the other two.

Diagram:

70μC...................48μC..............-80μC
O_____________O_____________O
...........0.35m.................0.35m

I began to understand it with two particles, but 3 completely lost me. I imagine 70, 48, and -80, would be q1, q2, and q3...d would be 0.35m. Would this be a problem where I use 1.6*10^-19?

Thanks :)

2. Jan 26, 2005

### neonerd

My answer for some reason, comes out completely different from what the book says. Could somebody please do out the work for the first particle?

Here is my work:

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

Answer in the book says: -1.4*10^2N

Last edited: Jan 26, 2005
3. Jan 26, 2005

### zitek

charge

q means charge.
1.6*10^19 C is the charge of an electron.
If there are 3 particles just do vector addition for the total force.

4. Jan 26, 2005

### neonerd

ah, I just realised my mistake :)

ke = 8.99 x 10 ^ 9 N/C ... not -9

that fixed it :)