Calculating Charge on Conducting Balls Hanging from Strings

In summary, two tiny conducting balls of identical mass but different charge hang from nonconducting threads of length 100 cm. Assuming that q is so small that tanq can be replaced by its approximate equal, sinq, x can be found by summing the forces acting on a charge: weight, Coulomb force, and tension.
  • #1
Gale
684
2
In the figure, two tiny conducting balls of identical mass m and identical charge q hang from nonconducting threads of length L. Assume that q is so small that tanq can be replaced by its approximate equal, sinq.

First find an expression for x in terms of L, q, m, k and g.

Then, given L = 100 cm, m = 7.00 g, and x = 3.00 cm, use the expresion above to calculate the magnitude of q: ((1.02e-8 C)) << correct answer

so i started with sumF=ma= kq^2/x^2 + Xcomp of the Tension.

I think I'm screwing up because i don't remember how tension works as a force.
 
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  • #2
Identify the forces acting on a charge: weight, Coulomb force, tension. Now apply equilibrium conditions for vertical and horizontal components.

(I assume the problem says to assume the angle the string makes with the vertical is small so that [itex]tan\theta = sin\theta = (x/2)/L[/itex]. I assume "x" is the distance between the two charges.)
 
  • #3
right i understood that much.
i got
kq/.003^2 + .1cos (theta)

i don't remeber what to put for the tension part of the eq.
 
  • #4
Call the tension T. It's an unknown; it will drop out of the equations.

Note: Don't start plugging in numbers until you've figured out the equation.
 
  • #5
Help

I too am working on a similar question. For mine it is asking for a proof that...
[tex]x=(\frac{2kLq^2}{mg})^{1/3}[/tex]

So far this is what I have
[tex]
\begin{equation*}
\begin{split}
F_{net}=0 \\
0=F_c+F_g+T \\
mg+\frac{kq^2}{x^2}+T=0
\end{split}
\end{equation*}
[/tex]

I'm not sure where to incorperate the [tex]\sin\Theta=\tan\Theta[/tex] Any more help would be more then amazing. Thanks Zac
 
  • #6
z_sharp said:
So far this is what I have
[tex]
\begin{equation*}
\begin{split}
F_{net}=0 \\
0=F_c+F_g+T \\
mg+\frac{kq^2}{x^2}+T=0
\end{split}
\end{equation*}
[/tex]
The three forces (Coulomb, tension, and weight) all act in different directions. So your last equation is incorrect. Instead, sum the horizontal and vertical components separately and set each sum equal to zero. (You'll find that things depend on the angle, which is where you can use a small angle approximation.)
 
  • #7
So following your advice this is what I have come up with but I can't seem to solve the proof

[tex]
\begin{equation*}
\begin{split}
F_{net_{x}}=0
\\ 0=F_c+T_x
\\ 0=\frac{kq^2}{L^2}-L\sin\theta
\end{split}\end{equation*}[/tex]

and

[tex]
\begin{equation*}
\begin{split}
F_{net_{y}}=0
\\ 0=F_g+T_y
\\ 0=mg-L\cos\theta
\end{split}\end{equation*}[/tex]

I'm not exactly sure where to go from there, I have tried setting the two equations equal to one another but can't seem to isolate the x properly. I have also tried subing in (x/2/L) for the sin Theta but that didn't really help me either. Any help would be more then great. Zac
 
  • #8
z_sharp said:
[tex]
\begin{equation*}
\begin{split}
F_{net_{x}}=0
\\ 0=F_c+T_x
\\ 0=\frac{kq^2}{L^2}-L\sin\theta
\end{split}\end{equation*}[/tex]
L is the length of the string, not the distance between the balls (call that x) or the tension (call that T). Rewrite this equation.
and

[tex]
\begin{equation*}
\begin{split}
F_{net_{y}}=0
\\ 0=F_g+T_y
\\ 0=mg-L\cos\theta
\end{split}\end{equation*}[/tex]
You'll need to rewrite this equation as well.

I'm not exactly sure where to go from there, I have tried setting the two equations equal to one another but can't seem to isolate the x properly. I have also tried subing in (x/2/L) for the sin Theta but that didn't really help me either.
Once you start with the correct equations, things will go smoother. :smile: Combine the two corrected equations: You'll get an expression for tan(theta). Then you'll be able to use the small angle approximation.
 

What are conducting balls on a string?

Conducting balls on a string are a simple scientific demonstration that shows the principles of electrical conduction. It involves two metal balls connected by a string, and when one ball is charged, the other ball also becomes charged, demonstrating the transfer of electrical charge.

How do conducting balls on a string work?

Conducting balls on a string work due to the principles of electrical conduction. When one ball is charged with electricity, it creates an electric field that affects the other ball, causing it to also become charged with the same type of electricity. The electric field is created by the movement of electrons from one ball to the other through the string.

What materials are needed to conduct this experiment?

To conduct the conducting balls on a string experiment, you will need two metal balls (such as aluminum or brass), a non-conductive string (such as nylon or cotton), and a source of electricity (such as a battery or static electricity generator).

What can conducting balls on a string demonstrate?

Conducting balls on a string can demonstrate the principles of electrical conduction and the transfer of electric charge. It can also show how objects with like charges repel each other, while objects with opposite charges attract each other.

What other experiments can be done with conducting balls on a string?

Conducting balls on a string can be used to conduct other experiments, such as demonstrating the transfer of charge through different materials, comparing the conductivity of different materials, or exploring the concept of static electricity. It can also be combined with other materials, such as balloons or rods, to create more complex demonstrations.

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