View Full Version : Coloumb's Law
TinTinTinTin
Nov 8, 2008, 04:53 PM
Hi, I am in grade 12 and in physics we are talking about Coulomb's law. I understand the equation Fe = kq1q2/r^2 but I am confused about the directions we assign.
For example if I have 3 spheres A, B, C in a straight line each having a charge of +2, -4, +6 respectively with a distance of 0.5m within each then what is the electrostatic force acting on C?
A B C
o ----------------------- o -------------------------- o
I know I am suppose to do forces with AC then BC then add them up. So for AC I would get a positive force. Would this be labelled Right because C is moving away? Or labelled left since A is moving away?
What about BC? I would get a negative force but how would I label it?
Thanks in advance and sorry if I wasn't clear
Capuchin
Nov 8, 2008, 05:06 PM
You're right, it's a little confusing because they both repel each other.
The best way is to label the force that the balls feel, so that's a force on A away from C and a force on C away from A, a force on B towards C and a force on C towards B. Does that help?
TinTinTinTin
Nov 8, 2008, 05:10 PM
Sorry, I am still confused
So if I do my calculation for AC (where A is acting on C) then I get the 1.08 x 10^11 N. Would I label this as Right since I am concerned with the electrostatic force on C and C in this case is repelled by sphere A?
Capuchin
Nov 8, 2008, 05:13 PM
Yes, you just want to label what you are interested in. In this case it's the force or movement of sphere C, so just label it as a force on C.
It's hard to answer your question, because the way that the teacher wants it labeled might not be the way that I would think to label it, I'm just trying to suggest something that's as clear as possible.
TinTinTinTin
Nov 8, 2008, 05:15 PM
All right thank you for your input, I might as well just ask my teacher then seeing there could be different ways to label the directions. Thanks again for your time
Capuchin
Nov 8, 2008, 05:19 PM
Oh, yes it would be right for the interaction from A and left for the interaction from B, then just add the vectors to get the net force.
You're welcome, feel free to stick around!
visharad
Nov 14, 2008, 03:06 AM
Use the following steps, then it will be very easy.
Suppose there are charges A and B. You need to find force by A on B.
Let r=distance between A and B,
q1=charge at A,
q2=charge at B
Calculate the force using formula kq1q2/r^2
Direction is from A to B.
Using above rule find all forces and do vector addition. If final result is negative, then take its absolute value and reverse the direction.
In your problem, force by A on C
= k*2*6/1.0^2
= 12k towards right
It is towards right because force is from A to C. A is on left and C is on right. So A to C becomes towards right.
Force by B on C = k*(-4)*6/0.5^2
= -96k towards right
So total force on C
= 12k - 96k = -84k towards right.
This means +84k towards left.
ngasnier
Nov 14, 2008, 05:02 AM
In regards to just your generally question about the sign (direction) of the electrical force:
Coloumb's law equation gives the unsigned MAGNITUDE of a force between two electrically charged objects.
Like charges always repel (negative force) and unlike charges attract (positive force). Sum them up for the total force.