A bead with charge q1 = 2.07 μC is fixed in place at the end of a wire which makes an angle of θ = 33.4° with respect to the horizontal. A second bead with mass m2 = 3.59 g and charge 4.30 μC slides without friction on the wire. What is the distance d where the force of the Earth's gravity on m2 is balanced by the electric force between the two beads? Neglect the gravitational interaction between the two beads.
I've set it up like this: m2 * 9.8 = (k * q1 * q2)/d^2) * sin(theta), but still it is incorrect. Any suggestions.
2006-09-02 08:13:42 · 3 answers · asked by eltel2910 1 in Science & Mathematics ➔ Physics
The component of the 2nd bead's weight that points along the wire must balance the force of opposition at the distance d. You have the term sin(theta) on the wrong side of the equation.
2006-09-02 09:33:56 · answer #1 · answered by sojsail 7 · 0⤊ 0⤋
Yes I think you have it about as correct as need be. Ask a scientist to half a number over and over and what is the answer? He will say you can't get there. Ask an engineer and he will say I can get close enough. Are you a scientist or engineer type?
2006-09-02 15:21:06 · answer #2 · answered by Benj 2 · 0⤊ 0⤋
poster above is correct, but also make sure you are using the correct trig function. I suspect it may be cos(theta), the wire makes angle theta w.r.t horizontal.
2006-09-02 19:44:35 · answer #3 · answered by tomz17 2 · 0⤊ 0⤋