Solving Dissipating Charge Physics Problem

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Homework Help Overview

The problem involves two small charged spheres suspended by threads, where the charge leaks off the spheres as they approach each other. The context is within electrostatics and dynamics, focusing on the forces acting on the spheres and their motion as they lose charge.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to analyze the forces acting on the spheres and differentiate the equations governing their motion. They express uncertainty about their approach and whether their assumptions, such as the approximation of angles, are valid.
  • Some participants suggest solving the original equation for charge and approximating angles, while others question the assumption of equilibrium in the system.

Discussion Status

The discussion is ongoing, with participants providing guidance on approximations and interpretations of the forces involved. There is an acknowledgment of errors in the original poster's approach, and some participants have offered alternative perspectives on the problem setup.

Contextual Notes

There are indications of missing information regarding the assumptions about equilibrium and the nature of the angles involved in the problem. The original poster expresses confusion about the complexity of the problem and the validity of their methods.

Saketh
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A friend of mine has given me a physics problem to solve. Here it is:
Two small equally charged spheres, each of mass m, are suspended from the same point by silk threads of length \ell. The distance between the spheres x << \ell. Find the rate \frac{dq}{dt} with which the charge leaks off each sphere if their approach velocity varies as v = \frac{a}{\sqrt{x}}, where a is a constant.​
I started by writing down the forces:
mg\tan{\theta} = \frac{q^2}{4\pi \epsilon_0 x^2}
Since x << \ell, I concluded that \sin{\theta} \approx \tan{\theta}. I rewrote the above force equation with sin instead of tan.

After this, I started doing things randomly. At first, I tried using angular speed with \ell as the radius of rotation, but that seemed unnecessarily complicated.

So I went back and differentiated both sides of the first equation. I don't know if this is correct, but I let \cos{\theta} be 1 since \theta \approx 0. So, using the quotient rule, I had:
mg = \frac{1}{2\pi \epsilon_0}\frac{q^2(\frac{dx}{dt}) - q\frac{dq}{dt}}{x^3}
At this point, since \frac{dx}{dt} = \frac{a}{\sqrt{x}}, I started substituting things in. However, I ended up with charge as a function of time, distance as a function of time, and the time derivative of the charge function all in one equation - unsolvable.

The other way that I tried it was to start from the premise that \ell \frac{d\theta}{dt} = \frac{a}{\sqrt{x}} and go from there. But I don't know if this is the correct way either.

Basically, I have no idea what I'm doing. My friend said this is an easy problem, but I am stumped. Am I approaching this incorrectly?

Thanks for the assistance.
 
Last edited:
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Solve your original equation for q and approximate \theta with x/l. Then differentiate q with respect to time. You should end up with dq/dt being constant within this approximation.

[Note: you may need to make minor adjustments to the preceding depending on whether your angle is the half-angle or full angle.]
 
I don't think one should assume that the spheres are in equilibrium. That is

T\sin(\theta) - F_Q = m\ddot{x}

since

v_x(x) \Rightarrow v_x(t)
 
Last edited:
Tide said:
Solve your original equation for q and approximate \theta with x/l. Then differentiate q with respect to time. You should end up with dq/dt being constant within this approximation.

[Note: you may need to make minor adjustments to the preceding depending on whether your angle is the half-angle or full angle.]
Thanks - I have found my errors now. I had to approximate \theta with \frac{x}{2\ell}.

For those who are interested, the answer is:
<br /> \frac{dq}{dt} = \frac{3}{2}a\sqrt{\frac{2\pi \epsilon_0 mg}{\ell}}<br />
 

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