- #1
verd
- 146
- 0
Hey,
I thought I did this correctly and my answer turned out incorrect. Here's the problem:
http://synthdriven.com/images/deletable/help.jpg"
This is what I did:
In order to use Coloumb's Law, I need r, the distance between the two charges.
[tex]\arcsin{4}=\frac{x}{0.24m}[/tex]
[tex]x=0.24\arcsin{4}=0.098m[/tex]
[tex]r=2x=0.196m[/tex]
So that's r.
Then I need to pick one side of this thing apart and break it down into x and y-components.
With the weight, I have three forces acting on this thing. The tension in the rope, the weight, and the electrical force repelling each sphere.
My goal is to find the x-component of the electrical force so I can then plug it into Coulomb's law and find the charge.
For the weight,
Wx=0
Wy=-mg=-0.98
For the tension,
Tx=-Tsin4
Ty=Tcos4
For the electrical force,
Fx=Tsin4
Fy=0
To find T, all I did was:
FyNET=Wy+Ty+Fy=Tcos4-mg
(net force of y-component)
T=mg/cos4
Right??
So for T, I get 0.098239
Plugging that into Tx, I get 0.006853And then it goes into Coulomb's law. Which I have reformatted this way:
[tex]F=\frac{kq^2}{r^2}[/tex]
...Because both charges are supposed to be equal... Moving that around, I get:
[tex]q=\sqrt{\frac{Fr^2}{k}}[/tex]
Plugging in values, I got 1.711e-7What did I do wrong?
I thought I did this correctly and my answer turned out incorrect. Here's the problem:
http://synthdriven.com/images/deletable/help.jpg"
This is what I did:
In order to use Coloumb's Law, I need r, the distance between the two charges.
[tex]\arcsin{4}=\frac{x}{0.24m}[/tex]
[tex]x=0.24\arcsin{4}=0.098m[/tex]
[tex]r=2x=0.196m[/tex]
So that's r.
Then I need to pick one side of this thing apart and break it down into x and y-components.
With the weight, I have three forces acting on this thing. The tension in the rope, the weight, and the electrical force repelling each sphere.
My goal is to find the x-component of the electrical force so I can then plug it into Coulomb's law and find the charge.
For the weight,
Wx=0
Wy=-mg=-0.98
For the tension,
Tx=-Tsin4
Ty=Tcos4
For the electrical force,
Fx=Tsin4
Fy=0
To find T, all I did was:
FyNET=Wy+Ty+Fy=Tcos4-mg
(net force of y-component)
T=mg/cos4
Right??
So for T, I get 0.098239
Plugging that into Tx, I get 0.006853And then it goes into Coulomb's law. Which I have reformatted this way:
[tex]F=\frac{kq^2}{r^2}[/tex]
...Because both charges are supposed to be equal... Moving that around, I get:
[tex]q=\sqrt{\frac{Fr^2}{k}}[/tex]
Plugging in values, I got 1.711e-7What did I do wrong?
Last edited by a moderator: