Question

A bead with a mass of 0.050 g and a charge of 20 nC is free to slide on a vertical rod. At the base of the rod is a fixed 20 nC charge. In equilibrium, at what height above the fixed charge does the bead rest?

Answer 1

Answer:

height above the fixed charge is 8.571 cm

Explanation:

given data

mass m = 0.050 g = 5 × $10^{-5}$ kg

charge bead q1 = 20 nC = 20 × $10^{-9}$ C

charge base q2 = 20 nC = 20× $10^{-9}$ C

to find out

what height above the fixed charge does the bead rest

solution

we know that when charge at rest then downward gravitational force is balance by electrostatic force so

$mg = k\frac{q1q2}{r^2}$   .............1

here k is 9 × $10^{9}$ Nm²/C² and g = 9.8 m/s² and r is height of bread  

put here all value in equation 1

$5*10^{-5}*9.8 = 9*10^{9} \frac{20*10^{-9}*20*10^{-9}}{(r^2}$

r² = 7.3469 × $10^{-3}$

r = 0.08571 m = 8.571 cm

so height above the fixed charge is 8.571 cm