Basic electrostatics problem; analytical solution?

[Taulant Sholla]

Homework Statement

For the system given, both objects have the same charge and same mass (both given). I'm also given string length, L. I need to solve for θ.

Homework Equations

Coulomb's Law, W=mg

The Attempt at a Solution

Using simple equilibrium force analysis (with weight, tension, and electrostatics forces), I get:

Is there a way to solve for θ analytically, or do I have to find a graphical solution?
Thank you!

 

[gneill]

It might be manipulated into the form of a cubic equation. Cubics have a closed form solution for their roots. It'll involve a couple of changes of variables I think. Try expressing cos(θ) in terms of sin(θ), and then call x = sin(θ) so you're working with x rather than trig.

 

[Ray Vickson]

Homework Statement

View attachment 112497
For the system given, both objects have the same charge and same mass (both given). I'm also given string length, L. I need to solve for θ.

Homework Equations

Coulomb's Law, W=mg

The Attempt at a Solution

Using simple equilibrium force analysis (with weight, tension, and electrostatics forces), I get:
View attachment 112499
Is there a way to solve for θ analytically, or do I have to find a graphical solution?
Thank you!

The equation $\sin^3(\theta)/\cos(\theta) = a$ is not difficult to solve analytically. For $a > 0$ we have $0 < \theta < \pi/2$, so $\cos(\theta) = \sqrt{1 - \sin^2(\theta)} > 0$. Therefore, the new variable $x = \sin^2(\theta)$ obeys the cubic equation $x^3/(1-x) = a^2$. The exact solution of this cubic is not too complicated or difficult to work with. From $x$ we can recover $\theta = \arcsin(\sqrt{x})$.

 

[andrevdh]

I get that
sin(θ) cos(θ) = kq²/4L²mg
?
which, if correct, then comes to

½ sin(2θ) = kq²/4L²mg

 

[Taulant Sholla]

I believe this is incorrect. I'm pretty sure of the solution I posted, since it does agree with computational results.

 

[Taulant Sholla]

Thank you so much!
I went down this path and saw the enormously complexity of solving cubics, which is - I guess - why we're taught to use graphical solutions methods instead.

The equation $\sin^3(\theta)/\cos(\theta) = a$ is not difficult to solve analytically. For $a > 0$ we have $0 < \theta < \pi/2$, so $\cos(\theta) = \sqrt{1 - \sin^2(\theta)} > 0$. Therefore, the new variable $x = \sin^2(\theta)$ obeys the cubic equation $x^3/(1-x) = a^2$. The exact solution of this cubic is not too complicated or difficult to work with. From $x$ we can recover $\theta = \arcsin(\sqrt{x})$.