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

height above the fixed charge is 8.571 cm

Step-by-step 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(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) (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