- #1
adkinje
- 11
- 0
Two point particles, each of mass m and charge q are suspended from a common point by threads of length L. Each thread makes an angle [tex]\theta[/tex] with the vertical. (I attached a diagram to help).
I must show that
[tex]q=2L\sin\theta\sqrt{\frac{mg}{k}\tan\theta}[/tex]
I start out by writing the force sum in the coordinate system, and then I solve for q:
[tex]0=mg+k\frac{q^2}{R^2}[/tex]
[tex]=mg+k\frac{q^2}{4L^2\sin^2\theta}[/tex]
[tex]q^2=-4L^2\sin^2\theta\frac{mg}{k}[/tex]
I'm confused by the negative sign, if I ignore it I get:
[tex]q=2L\sin\theta\sqrt{\frac{mg}{k}}[/tex]
The [tex]\tan\theta[/tex] is missing in my solution.
I must show that
[tex]q=2L\sin\theta\sqrt{\frac{mg}{k}\tan\theta}[/tex]
I start out by writing the force sum in the coordinate system, and then I solve for q:
[tex]0=mg+k\frac{q^2}{R^2}[/tex]
[tex]=mg+k\frac{q^2}{4L^2\sin^2\theta}[/tex]
[tex]q^2=-4L^2\sin^2\theta\frac{mg}{k}[/tex]
I'm confused by the negative sign, if I ignore it I get:
[tex]q=2L\sin\theta\sqrt{\frac{mg}{k}}[/tex]
The [tex]\tan\theta[/tex] is missing in my solution.