# Coulomb's Law and Tension

1. Jan 23, 2012

### dancer2012

1. The problem statement, all variables and given/known data

Two small pith balls, each of mass m = 12 g, are suspended by 1.2 m fine (so that we can neglect their mass in this problem) strings and are not moving. If the angle that each string makes with the vertical is θ = 42.6°, and the charges on the two balls are equal, what is that charge (in μC)?

2. Relevant equations

F(T)+F(g)+F(E)=0
F(Ty)=F(g)=mg
F(Tx)=k(q^2/r^2)

3. The attempt at a solution

I really don't know how to manipulate these formulas to find an answer. I know you find F(Ty) first, but then I don't know how to get F(Tx). I think I am missing a formula somewhere. Please Help Thank you

2. Jan 23, 2012

### Mindscrape

You're pretty close, but the problem has an additional piece of information that you haven't used yet. Hint: it's the angle!

3. Jan 23, 2012

### Simon Bridge

Probably best to draw free-body diagrams first perhaps?
Hi Dancer, welcome to PF.

4. Jan 24, 2012

### dancer2012

The force diagram is attached

F(Ty)=cos(θ)*F(T) - - - F(T)=F(Ty)/cos(θ)
F(Tx)=sin(θ)*F(T) - - - F(T)=F(Tx)/sin(θ)

So, F(Ty)/cos(θ)=F(Tx)/sin(θ) - - - F(Tx)=[F(Ty)*sin(θ)]/cos(θ) - - - F(Tx)=F(Ty)*tan(θ)

Did I do all of that correctly?

5. Jan 24, 2012

### dancer2012

Sorry here is the force diagram actually

#### Attached Files:

• ###### Force Diagram.jpg
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6. Jan 24, 2012

### Simon Bridge

The next step after drawing the diagram is to add up the forces in each direction. Start with just the left-hand ball. eg. $$F_T\cos\theta=F_g$$... do the horizontal one, then write out what each of the forces actually are. eg. $$F_e=\frac{kq^2}{r^2}$$... and r is related to the angle... see?

7. Jan 24, 2012

### dancer2012

I understand that

8. Jan 24, 2012

### Mindscrape

Yeah, that's right. tanØ=Tx/Ty

So you're set, work out the algebra from here. Don't forget, as Simon mentioned, to get the EM force in terms of "x distance."

9. Jan 25, 2012

### dancer2012

F(Tx)=mg*tan(θ)

q=√((F(Tx)*r^2)/k)

would r be mg/tan(θ)??

10. Jan 25, 2012

### Simon Bridge

Complete:
if the pith balls were at positions (±x,-y) ...
then there is a triangle l:y:x with angle θ between l and y (l=length of the string)
From this picture:

the distance between the balls is ____________.

and tanθ=_x/y_, sinθ=_____, and cosθ=_____.

Last edited: Jan 25, 2012
11. Jan 25, 2012

### Mindscrape

Nah, that would be the horizontal force, which is the coulomb force you're in part trying to solve for. You've a string length you haven't used yet. ;)

12. Jan 26, 2012

### dancer2012

Ok so sin(θ)*string length=r

SO..
F(Ty)=F(g)=mg
F(Tx)=k(q^2/r^2)
F(Tx)=F(Ty)*tan(θ)
q=√((F(Tx)*r^2)/k)

F(Tx)=mg*tan(θ) - - - - - - - - - F(Tx)=.012kg*9.81m/s^2=.11772N→A
q=√((F(Tx)*(sin(θ)*L)^2)/k)----q=√(A*(sin(42.6)*1.2)^2/9E9)=2.938E-6C or 2.9μC

Is 2.9 μC the correct answer or did I make a mistake?

13. Jan 26, 2012

### Mindscrape

Careful, you want the total distance between the pith balls. You just found one half, one leg.