A rigid, insulating fiber runs along a portion of the y-axis; the fiber isnot free to move. Gravity acts downward (g = 9.81 m/s2).A charge Qa = -7 µC is fixed to the fiber at the origin. A bead with a hole drilled through its center is slipped over the fiber andis free to move along the fiber without friction. The mass of the bead is m = 250g and its charge is Qb. At equilibrium, the bead floats a distanceyb = 11 cm above the origin.
Calculate the charge on the bead.
Why isn't coulonmbs law working for me????
The Attempt at a Solution
OK heres what i did
F = (KQa * Qb) / r^2
(absolute values for the charges)
(i got F by realizing that it must balance the opposing downward force (drew FBD)
(.245)9.81 = (9.0*10^9) (7*10^-6)Qb / ((.011)^2)
So Qb comes to 4.71*10-9
=wrong for sum reason
Please help and THANKS!