Question

A positively charged particle Q1 = +45 nC is held fixed at the origin. A second charge Q2 of mass m = 4.5 μg is floating a distance d = 25 cm above charge The net force on Q2 is equal to zero. You may assume this system is close to the surface of the Earth.
|Q2| = m g d2/( k Q1 )
Calculate the magnitude of Q2 in units of nanocoulombs.

Answer 1

Answer:

( About ) 6.8nC

Explanation:

We are given the equation |Q2| = mgd^2 / kQ1. Let us substitute known values into this equation, but first list the given,

Charge Q2 = +45nC = (45 × 10⁻⁹) C

mass of charge Q2 = 4.5 μg, force of gravity = 4.5 μg × 9.8 m/s² = ( 4.41 × 10^-5 ) N,

Distance between charges = 25 cm = 0.25 m,

k = Coulomb's constant = 9 × 10^9

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And of course, we have to solve for the magnitude of Q2, represented by the charge magnitude of the charge on Q2 -

(4.41 × 10^-5) = [(9.0 × 10⁹) × (45 × 10⁻⁹) × Q₂] / 0.25²

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Solution = ( About ) 6.8nC