Calculating Charge on Conducting Balls Hanging from Strings

[Gale]

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.

 

[Doc Al]

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 $tan\theta = sin\theta = (x/2)/L$. I assume "x" is the distance between the two charges.)

 

[Gale]

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.

 

[Doc Al]

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.

 

[z_sharp]

Help

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

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

I'm not sure where to incorperate the $$\sin\Theta=\tan\Theta$$ Any more help would be more then amazing. Thanks Zac

 

[Doc Al]

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

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.)

 

[z_sharp]

So following your advice this is what I have come up with but I can't seem to solve the proof

$$\begin{equation*}<br /> \begin{split}<br /> F_{net_{x}}=0 <br /> \\ 0=F_c+T_x <br /> \\ 0=\frac{kq^2}{L^2}-L\sin\theta<br /> \end{split}\end{equation*}$$

and

$$\begin{equation*}<br /> \begin{split}<br /> F_{net_{y}}=0 <br /> \\ 0=F_g+T_y <br /> \\ 0=mg-L\cos\theta<br /> \end{split}\end{equation*}$$

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

 

[Doc Al]

$$\begin{equation*}<br /> \begin{split}<br /> F_{net_{x}}=0 <br /> \\ 0=F_c+T_x <br /> \\ 0=\frac{kq^2}{L^2}-L\sin\theta<br /> \end{split}\end{equation*}$$

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

$$\begin{equation*}<br /> \begin{split}<br /> F_{net_{y}}=0 <br /> \\ 0=F_g+T_y <br /> \\ 0=mg-L\cos\theta<br /> \end{split}\end{equation*}$$

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. Combine the two corrected equations: You'll get an expression for tan(theta). Then you'll be able to use the small angle approximation.