Coulomb's Law lab , , it do tommorrow

[Suy]

Coulomb's Law lab , please help, it do tomorrow

Homework Statement

Here is the picturehttp://img407.imageshack.us/img407/933/48460315.jpg
Distance(cm) radius(cm)
0.2 1.4
0.4 1.4
0.8 1.35
1.0 1.35
1.2 1.3

This is the procedure:
Set up apparatus. Charge the sphere on the block by induction. Observe what happens to the suspended sphere as the charged sphere is brought close to it. Allow the suspended sphere to come into contact with the first.
Allow suspended sphere to return to rest position. Bring the furst sphere towards the suspended sphere again to make mesurements.

Homework Equations

F_e=mg*d/L
this is the equation that teacher gave us, but i don't understand this equation at all.

The Attempt at a Solution

I don't really understand the equation F_e=mg*d/L, why is it d/L(distance/length of string, right)
and why we don't use F_e=(kq₁q₂)/r²
Also, I need 3 source of error , not including human error,
but i have no clue what is it, it do tomorrow , if anyone can help me///

 

[rl.bhat]

Resultant of mg and Fe makes an angle θ with the vertical.
Draw the figure and find out what is d/L stands for and how it is related to Fe and mg.

 

[Suy]

Resultant of mg and Fe makes an angle θ with the vertical.
Draw the figure and find out what is d/L stands for and how it is related to Fe and mg.

I still don't get it...

 

[rl.bhat]

If Fx and Fy are the components of a force F, and if θ is the angle between F and Fx, then tanθ = ?.

 

[Suy]

is it right?, but how do u find y

 

[rl.bhat]

View attachment 20684
is it right?, but how do u find y

I can't see the attachment.
If θ is very small, tanθ is nearly equal to θ, and d/L is θ.

 

[Suy]

http://img200.imageshack.us/img200/3631/34175284.jpg
do you mean tanθ=(y/x)

 

[rl.bhat]

No. The resultant of Fe and mg will be along the string of length L. In this position string makes an angle θ with the vertical and the suspended sphere displaced a small distance d, then tanθ = d/L = Fe/mg.

 

[Suy]

Can u draw a picture for me, i am confused...
sorry, but i really don't get it..

 

[rl.bhat]

Sorry. I don,t know how to draw the picture. Anyway I will explain from your picture.
You have shown the separation of two positions of the suspended charged sphere as x or d.
In the equilibrium position of the sphere, F acts horizontally and mg acts vertically.
The relation between the length of the arc d, radius L and subtended angle θ is d = Lθ.

 

[Suy]

Fnially, thanks a lot,
now i understand what you mean,
since tanθ=F/(mg) and tanθ=d/(L), so F/(mg)=d/L, if i solve for F, F=mg*(d/L)
but i don't understand why do we use F_{e x-component
http://img200.imageshack.us/img200/3631/341752}

 

[rl.bhat]

Where are we using Fe x-component?

 

[Suy]

http://img200.imageshack.us/img200/3631/34175284.jpg
this picture, i guess i am wrong..

 

[Suy]

i have 2 more question, the teacher asked to graph force vs r and straighten the graph,
Do i use 1/r2 ?
and F=mg*(d/L),
g=9.81m/s, or -9.81m/s?

a point charge of -0.35 nC is fixed at the origin. where must and electron be placed in order for the electric force acting on it to be exactly opposite to its weight?
To balance the force, mg=k(q1q2)/r2
q1=-0.35 nC
but i am not sure how to find q2 and the mass

 

[rl.bhat]

I don't understand what he means by straighten the graph.
g = 9.8 m/s^2

You have that the weight of the electron is equal to the force between q1 and q2.
Mass of the electron and its charge is given in any book.

 

[Suy]

I think he wanted us to make a straight line, not a curve
i remember the relation between F and r is
The gravitation attraction force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of their separation distance.
is this same as electrostatics,because they have similar equation

 

[MattH25]

to straighten your curve you have to use the inverse of radius sqaured (1/r2)