An electric point charge of Q1 = 9.27 nC is placed at the origin of the real axi

Question

An electric point charge of Q1 = 9.27 nC is placed at the origin of the real axis. Another point charge of Q2 = 1.01 nC is placed at a position of p = 2.39 m on the real axis. At which position can a third point charge of q = -7.18 nC be placed so that the net electrostatic force on it is zero?

That answer is 1.797m The question I cannot get is the second part which is....


Let the sign of Q2 be changed from positive to negative. At which position can the point charge q be placed now so that the net electrostatic force on it is zero?

Explanation / Answer

As charges Q1,Q2 are having same sign, Charge Q3 should be kept in between these two charges on the line joining them.
Force on charge q due to Q1 = k*Q1*q/r1^2
Force on charge q due to Q2 = k*Q2*q/(2.39-r1)^2
Hence r2 = 2.39 - r1
k*Q1*q/r1^2 = k*Q2*q/(2.39-r1)^2
Q1/r1^2 = Q1/(2.39-r1)^2
(9.27*10^(-9))/r1^2 = (1.01*10^(-9))/(2.39-r1)^2
r1/(2.39-r1) = 3.03
r1 = (3.03*2.39)/4.03 = 1.8 m
r2 = 2.39 - 1.8 = 0.59 m

b) Let charge q is at distance r away from both the charges on the line joining Q1&Q2 .

r2 = r

r1 = 2.39 + r

k*Q1*q/r1^2 = k*Q2*q/(2.39-r1)^2

(9.27*10^(-9))/(2.39 + r)^2 = (1.01*10^(-9))/r^2

(2.39 + r)/r = 3.03

2.03 r = 2.39

r = 1.177 m

Hence r2 = 1.777 m

r1 = 2.39 + 1.777 = 3.567 m

Hence, charge q should be at a distance 1.777m from Q2 and 3.567 m from Q1 on the line joining Q1&Q2 .

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