A be ad with charg e q 1 = 1.07 ?C is fixed in place at the end of a wire which

Question

A bead with charge q1 = 1.07 ?C is fixed in place at the end of a wire which makes an angle of ? = 33.6

A bead with charge q1 = 1.07 ?C is fixed in place at the end of a wire which makes an angle of ? = 33.6 degree with respect to the horizontal. A second bead with mass m2 = 5.55 g and charge 6.85 ?C slides without friction on the wire. What is the distance d where the force of the Earth's gravity onm2 is balanced by the electrostatic force between the two beads? Neglect the gravitational interaction between the two beads.

Explanation / Answer

because total force on q2 = 0 so m2*g*sin33.6 = kq1q2/d^2

to solve this eqation we get d =1.471m

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