Answer:
q(a) = 15 nC
q(b) = 10 nC
Explanation:
The distance between q(a) and q(b) is 5 cm
Also, q(a) + q(b) = 29 nC
Force between them, F(a,b) = 5.4*10^-4 N
Using Coulomb's law
F = kq1q2 / r², where
F = force between the charges
K = constant of proportionality
q1 = magnitude of first charge
q2 = magnitude of the second charge
r = distance between the charges
5.4*10^-4 = {8.99*10^9 * q(a)[25 - q(a)] *10^-18} / 0.05²
5.4*10^-4 = {8.99*10^9 * q(a)[25 - q(a)] *10^-18} / 0.0025
{8.99*10^9 * q(a)[25 - q(a)] *10^-18} = 5.4*10^-4 * 0.0025
{8.99*10^9 * q(a)[25 - q(a)] *10^-18} = 1.35*10^-6
q(a)[25 - q(a)] *10^-18 = 1.36*10^-6 / 8.99*10^9
q(a)[25 - q(a)] *10^-18 = 1.50*10^-16
q(a)[25 - q(a)] = 1.50*10^-16 * 10^18
q(a)[25 - q(a)] = 150
25q(a) - q(a)² = 150
q(a)² - 25q(a) + 150 = 0 solve using quadratic equation, we have
q(a) - 10 = 0
q(a) - 15 = 0
Since the question stated that q(a) had the higher charge, then q(a) = 15 nC and q(b) = 10 nC
