Equilateral triangle coulombs law problem

  • #1
[

Homework Statement


1.) Three charges are placed 25 cm apart in an equilateral triangle. Q1 has a charge of -6.0 nC, q2 has a charge of 4.0 nC and q3 has a charge of 4.0 nC. what is the force on q2?

q1 is the top point, q2 is the bottom left leaving q3 on the bottom right


Homework Equations



F = K q1q2/r^2

The Attempt at a Solution



Fnet2 = F12 + F32

using the above equation for f12 i get: 3.45 * 10^-6

components:(3.45 * 10^-6) sin60 = 2.99 * 10^-6 North
:(3.45 * 10^-6) cos60 = 1.72*10^-6 East

F32 using the equation above = 2.3 *10^-7 West

combining the x components = 5.81 *10^-6 West

so using pythagoras: Root((5.81 * 10^-6)^2 + (2.99*10^-6)^2))

= 3.05 * 10 ^-6 C

angle= arctan((2.99*10^-6/ 5.81*10^-6))
= 79 degrees North of west

so my final answer is 3.05 mirco coulombs (79 degrees N of W)

Is this correct? I want to make sure i have a grasp on these problems before my final exam on the 18. Any help is greatly appreciated!
 

Answers and Replies

  • #2
i used 10^-9 and meters, and i see i made a mistake in the answer it should be 3.05 newtons not micro coulombs.
 
  • #3
70
0
OK firstly I'd say that drawing a quick diagram really helps with visualising a problem like this. Here's one if you don't have one already http://yfrog.com/e4pf3ij

:smile:

Using that equation:

[tex]F=\frac{k.q_{1}.q_{2}}{r^{2}}[/tex]

You can indeed work out each of the forces, i.e. of each charge on each other charge, as you've done already.

So..

[tex]F_{1 on 2} = \frac{k\times (-6)\times {4}}{(25)^{2}} = \frac{-24k}{625}[/tex]

Then you need to account for the angle between them (since not on a straight line) so should actually be:

[tex]F_{1 on 2} = \frac{-24k}{625}sin(30)[/itex]

Then you can use the same method to calculate [itex]F_{3 on 2}[/itex] noting that the angle between them is 0 i.e. don't need to account for an angle between these.

Hopefully this helps a bit more :smile:
 
  • #4
70
0
Just to add that you should find that [itex]F_{2 on 3}=F_{3 on 2}[/itex] so that you only need to worry about the components [itex]F_{1 on 2}[/itex] and [itex]F_{2 on 1}[/itex].

Also you can simplify the calculations using that [itex]sin(30)+sin(60)=\frac{1+\sqrt{3}}{2}[/itex] and that [itex]\frac{kq_{1}q_{2}}{r^{2}}[/itex] components of both are equal.

Hence you should be able to work out the resulting force on charge [itex]q_{2}[/itex]
 
  • #5
Just to add that you should find that [itex]F_{2 on 3}=F_{3 on 2}[/itex] so that you only need to worry about the components [itex]F_{1 on 2}[/itex] and [itex]F_{2 on 1}[/itex].

Also you can simplify the calculations using that [itex]sin(30)+sin(60)=\frac{1+\sqrt{3}}{2}[/itex] and that [itex]\frac{kq_{1}q_{2}}{r^{2}}[/itex] components of both are equal.

Hence you should be able to work out the resulting force on charge [itex]q_{2}[/itex]
okay im slightly confused, why did you use 30 degrees instead of 60? and is my overall answer correct?
 
  • #6
70
0
Ok..

using the above equation for f12 i get: 3.45 * 10^-6
Shouldn't it be:

[tex]F_{1 on 2}=\frac{k\times q_{1}\times q_{2}}{r^{2}}=\frac{(9\times10^{9}N.m^{2}.C^{-2})(-6.0\times 10^{-9}C)(4.0\times 10^{-9}C)}{(25m)^{2}}=-3.45\times 10^{-10}N[/tex]

Then account for the angle between them, so:

[tex]F_{1 on 2}=-3.45\times 10^{-10}sin(30)N = -1.73\times 10^{-10}N[/tex]

Also, since [itex]F_{3,2}=F_{2,3}[/itex] then these forces just cancel each other out right.

Then the other force to consider:

[tex]F_{2 on 1}=\frac{k\times q_{2}\times q_{1}}{r^{2}}=\frac{(9\times10^{9}N.m^{2}.C^{-2})(4.0\times 10^{-9}C)(-6.0\times 10^{-9}C)}{(25m)^{2}}=-3.45\times 10^{-10}N[/tex]

Then account for the angle:

[tex]F_{2 on 1}=-3.45\times 10^{-10}sin(60)N = -2.99\times 10^{-10}N[/tex]

Therefore the force on [itex]q_{2}[/itex] is:

[tex]F=4.6\times 10^{-11}N[/tex]

Well, that's what I've just got anyway. Your whole method is a bit different. Hopefully that's clearer now :smile: (for both of us, was getting bit confused as well :wink:)
 

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