Two spheres hanging, find charge.

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Two identical spheres, each with a mass of 2.0g, are suspended from a central point by a 0.6m insulating line, forming a 30° angle due to identical electric charges. The problem involves calculating the charge on each sphere using Coulomb's Law and the forces acting on the spheres. The vertical force due to gravity and the horizontal repulsive force must be balanced, leading to the equation Fe = mg * tan(15°). After discussions and calculations, the correct charge is determined to be approximately 1.19 x 10^-7 C, highlighting the importance of careful force analysis and trigonometric relationships in solving the problem. The conversation emphasizes the need to verify published answers against calculated results.
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Homework Statement



Two identical, small spheres of mass 2.0g are fastened to the ends of a 0.6m long light, flexible, insulating fishline. The fishline is suspended by a hook in the ceiling at it's exact center. The spheres are each given the same electric charge. They are in static equilibrium, creating an angle of 30° between the halves of the string. Calculate the magnitude of the charge on each sphere.

Here's a sketch of the picture given on the worksheet:
Owokl.gif

Red lines are what I drew over the diagram to make a right triangle.

This being a bonus type question, I was given three possible answers and had to find out which one was the real one:

q =
3.7 x 10^-7 C or
1.2 x 10^3 C or
1.2 x 10^-3 C.

Homework Equations



Coulombs Law: F = k \ \frac{q_1 q_2}{r^2}

The Attempt at a Solution



First I converted the mass into kilograms: m = 0.002kg

Then I split the angle into two right triangles to get the distance between the spheres.

r/2 = Sin15 x 0.3 m
r = approx. 0.16m

Then I figured since the spheres are at rest, then a = 0 and Fg = Fe...so I tried coulombs law to solve for q:

F_{g} = F_{e}
mg = k \ \frac{q_1 q_2}{r^2}
\frac{mgr^2}{k} = q_1 q_2

I ended up with 2.6 x 10^-11...obviously way off any of the true answers.

Thanks for taking the time to read over all this. =D

Cheers - Krunklehorn
 
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You need to look at the forces acting on the balls.
There is a vertical force = weight of a ball = 2x10^-3x9.81 = 0.0196N
There is a sideways force, F, due to the repulsion.
Can you see that F/0.0196 = Tan15
This will give you F to use in the repulsion equation.
I got Q = 1.2x10^-7C, on your list there is 1.2x10^-3C
I use k = 8.98x10^9 in the repulsion equation
 
technician said:
You need to look at the forces acting on the balls.
There is a vertical force = weight of a ball = 2x10^-3x9.81 = 0.0196N
There is a sideways force, F, due to the repulsion.
Can you see that F/0.0196 = Tan15
This will give you F to use in the repulsion equation.
I got Q = 1.2x10^-7C, on your list there is 1.2x10^-3C
I use k = 8.98x10^9 in the repulsion equation

Oh I see! Fe pushes the spheres to the side while Fg pushes them down to the ground.

Tan = Opp / Adj so my equation should be:

Tan15 Fg = Fe

\frac{Tan15mgr^2}{2k} = q

Awww too bad it's actually wrong. I am getting 7.03x10^-15 as an answer.
 
Agree with technician. The force on one of the spheres is equal to the x-component of the fishline force(S):

Sx=S*sin15

and

S=Sy/cos15

Therefore

q=1.19*10-7

since

Fe=m*g*tan15
 
Last edited:
nure said:
since

Fe=m*g*tan15

Oh I completely agree with him and appreciate the help immensely...I just don't see why I'm getting such a mangled answer.

Did I rearrange the equation wrong?
 
The equation

Fe=(k*q2)/(r2)

can(as you know) be rewritten as

q=r*Sqrt[Fe/k]

Sum of forces in x-direction equals zero:

Fe-Sx=0

therefore

Fe=Sx

Similarly in the y-direction:

Sy-w=0

therefore

Sy=w=m*g

Now, from simple trigonometry you see that

Sy=S*cos15

and

Sx=S*sin15

This means that you can write Sx as

Sx=(Sy/cos15)*sin15=m*g*tan15

And from the equilibrium equation in the x-direction you see that

Fe=m*g*tan15

Now you have Fe. You then need the distance r between the spheres:

r=2*0.3*sin15

Filling all this information into the expression for q then gives you

q=1.187*10-7 C
 
Last edited:
Thanks for the help! I understand completely now.
 
Great...well done...don't always believe published answers
 
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