Coulomb's Law - Calculating Net Force

[Valentine]

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:

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.

 

[mfb]

9 is just a rounded version of 8.99, and the minus in the exponent was a mistake.

I am not understanding why the q1 and q2 are (70*10^-6) (-80*10^-6)

with Coulomb there. The charges are defined to have those values in the problem statement (µC, "micro-coulomb", means 10^-6 Coulomb).
If you want to calculate the force between the 70µC and the -80µC charge: the distance between the charges is not 0.35m there.

 

[haruspex]

says that meters and C cancel out

They only cancel out if all the distances plugged in are in metres and all the charges plugged in are in Coulombs.

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

That's not strictly correct. Since F, q1, q2 and r are represented symbolically at this point, their units are unspecified in the equation. Therefore the units of the constant should be shown:
F=(9x10⁹) q1 q2 / r² Nm²/C²
When the values are plugged in:
F=(9x10⁹) (70μC)*(48μC) / r² (Nm²/C²)
=(9x10⁹) 70*(48) / r² (Nm²/C²) μC²
= (9x10⁹) 70*(48) / r² Nm² (μC/C)²
= (9x10⁹) 70*(48) / r² Nm² (10^-6)²
= (9x10^-3) 70*(48) / r² Nm²
= (9x10^-3) 70*(48) / (0.35m)² Nm²
= (9x10^-3) 70*(48) / (0.35)² N

q1 and q2 are (70*10^-6) (-80*10^-6) and how neonerd got (2.46x10^2N)

neonerd got that number for the case where the charges are 48 and 70 μC.

 

[Valentine]

with Coulomb there. The charges are defined to have those values in the problem statement (µC, "micro-coulomb", means 10-6 Coulomb).
If you want to calculate the force between the 70µC and the -80µC charge: the distance between the charges is not 0.35m there.

Micro! I feel very slow for not realizing it was micro That explains why it was 10^-6. Thank you! Also, I would have to use 0.7m^2 correct if it was from q1 to q3?
I am trying the math and my numbers are not adding up. I am usually great with math but -- well I'll just show my work.

That's not strictly correct. Since F, q1, q2 and r are represented symbolically at this point, their units are unspecified in the equation.

I see! I had just crossed out mine but I think practicing it the way you're showing would be better for the future because I am sure it will come up and try to haunt me. I'll be ready for it though!

F= K q1 q2 / r^2
=(9*10^9N*m^2/C^2) (70*10^-6C) (48x10^-6C) / 0.32^2 (0.35^2 Because it is from q1 to q2)
= I ended up canceling out the C^2 and m^2 even though I understand haruspex's work and ended up with:
(9x10^-3N) (70) (48) / 0.35^2 which is similar to haruspex's work but I left the N in the K area.
=Multiplying the top I got 30.24N/0.35^2
=Finally I get 246.857

I thought that it was my signs and double check and triple checked but I am not getting -1.4x10^2 or any numbers around 140.

________________

Okay I am seeing why. I am needing to come up with k q1 q3/r^2 THEN add them (I think)

________________

Yes! After doing the equation for q1 q3 I added up the two and got the answer! Thanks mfb and haruspex for your patience and time! I appreciate it now to do the other two parts of this problem! :D