A bead with charge q 1 = 1.12 ?C is fixed in place at the end of a wire which ma

Question

A bead with charge q1 = 1.12 ?C is fixed in place at the end of a wire which makes an angle of ? = 30.6

A bead with charge q1 = 1.12 ?C is fixed in place at the end of a wire which makes an angle of ? = 30.6 degree with respect to the horizontal. A second bead with mass m2 = 4.54 g and charge 6.04 ?C 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 electrostatic force between the two beads? Neglect the gravitational interaction between the two beads.

Explanation / Answer

Charges q 1 = 1.12 x 10 -6 C q 2 = 6.04x 10 -6 C m 2 = 4.54x 10 -3 kg Angle ? = 30.6 o If the repulsive force between two charges equal to component of weight of mass m 2 i.e., K q 1 q 2 / d 2 = m 2 g sin ? From this d = ?[K q 1 q 2 / m 2 g sin ?] Where K = Coulomb's constant = 8.99 x 10 9 Nm 2/ C 2 Substitute values we get d = ?[2.682] =1.63782817306m

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