If one charge attracts the other with a force of 0.138N, what are the magnitudes of the two charges if their total charge is also 18.8 ?C? The charges are at a distance of 1.30 m apart. Note that you may need to solve a quadratic equation to reach your answer.
Given that F = 0.138 N
Q1+Q2 = 18.8*10^-6 C..(1)
F = k*Q1*Q2/r^2 = 0.138
Q1*Q2 = 0.138*r^2/k = (0.138*1.3^2)/(9*10^9) = 25.9*10^-12
Q1-Q2 = sqrt((Q1+Q2)^2-4*(Q1*Q2)) = sqrt((18.8*10^-6)^2-(4*25.9*10^-12)) = 15.8*10^-6.........(2)
solving (1) and(2) we get
2*Q1 = (18.8+15.8)*10^-6
Q1 = 17.3 uC
Q2 = 1.5 uC
Q1+Q2 = 18.8e-6 C
Q2 = 18.8e-6 - Q1 .............(1)
k*Q1*Q2/r^2 = F
(9e9*Q1*Q2)/(1.3*1.3) = 0.138
Q1*Q2 = 2.59e-11 ........(2)
substitutin 1 in 2
Q1*(18.8e-6-Q1) = 2.59e-11
Q1^2 - 18.8*10^-6Q1 + 2.59*10^-11 = 0
Q1 = 1.49*10^-6 C <-----answer
Q2 = 18.8*10^-6 - 1.49*10^-6
Q2 = 17.31*10^-6 C <--answer
Given,
Force of attraction of two charges = 0.138N
Total chare = 18.8 C and distance = r = 1.30m
Coulomb
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