adkinje
Jul25-09, 06:42 AM
Two point particles, each of mass m and charge q are suspended from a common point by threads of lenght L. Each thread makes an angle \theta with the vertical. (I attached a diagram to help).
I must show that
q=2L\sin\theta\sqrt{\frac{mg}{k}\tan\theta}
I start out by writing the force sum in the coordinate system, and then I solve for q:
0=mg+k\frac{q^2}{R^2}
=mg+k\frac{q^2}{4L^2\sin^2\theta}
q^2=-4L^2\sin^2\theta\frac{mg}{k}
I'm confused by the negative sign, if I ignore it I get:
q=2L\sin\theta\sqrt{\frac{mg}{k}}
The \tan\theta is missing in my solution.
I must show that
q=2L\sin\theta\sqrt{\frac{mg}{k}\tan\theta}
I start out by writing the force sum in the coordinate system, and then I solve for q:
0=mg+k\frac{q^2}{R^2}
=mg+k\frac{q^2}{4L^2\sin^2\theta}
q^2=-4L^2\sin^2\theta\frac{mg}{k}
I'm confused by the negative sign, if I ignore it I get:
q=2L\sin\theta\sqrt{\frac{mg}{k}}
The \tan\theta is missing in my solution.