Question

A positively charged particle Q1 = +15nC is held fixed at the origin. A second charge Q2 of mass m = 9.5g is floating a distance d = 25cm above charge Q1. The net force on Q2 is equal to zero. You may assume this system is close to the surface of the Earth.
Calculate the magnitude of Q2 in units of nanocoulombs.

Answer 1

Answer:

$Q_2=4.4293\times 10^{22}\ nC$

Explanation:

Given

charge on the fixed particle, $Q_1=15\times 10^{-9}\ C$

mass of the second charge, $m_2=9.5\times 10^{-3}\ kg$

distance between the fixed charge and the floating charge on the top, $d=0.25\ m$

• According to the question the second charge is floating just above the fixed positive charge despite of gravity this means that the floating charge is also positive in nature and hence feels the repulsion from the fixed charge which is equal in magnitude to the gravitational force on the charge.

$m_2.g=\frac{1}{4\pi\epsilon_0} \times \frac{Q_1.Q_2}{d^2}$

where:

$\epsilon_0=$ permittivity of free space

g = acceleration due to gravity

$9.5\times 10^{-3}\times 9.8=8.85\times 10^{-9}\times \frac{15\times 10^{-9}\times Q_2}{0.25^2}$

$Q_2=4.4293\times 10^{22}\ nC$