Question

Two point particles separated by 0.8 m carry a total charge of 185

Answer 1

let q1 and q2 be the charges

so givne q1+q2 = 185 uC-------------------1

apply FOrce F = kq1q2/r^2

as force is repulsier, both must be same charges(sign)

so

q1q2 = Fr^2/K

so

q1q2 = 80 * 0.8*0.8/9e9

q1q2 = 56.88*10^-10

now use the identitiy (q1-q2)^2 = (q1+q2)^2 - 4q1q2

(q1-q2)^2 = (185e-6*185e-6) -(4* 56.88 e-10)

q1-q2)^2 = 1.1473 e-8

q1-q2 = 107.11 uC ---------------------------2

adding 1 and 2

2q1 = 292.11uC

so q1 = 146 uC ------<<<<<<<<<<<<<(larger charge)

q2 = 185-146 = 39uC --------<<<<<< (SMALLER CHARGE)

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now if Force is attractive, bioth charges will have opposite sign

so F = - kq1q2/r^2

q1q2 = -56.88 e -10

(q1-q2)^2 = ((185e-6*185e-6) +(4* 56.88 e-10)

q1-q2 = 238.69 uC -----------------------------3

adding 1 and 3

2q1 = 423.69 uC

q1 = 211.84 uC ------------------<<<<<<<<<<<larger charge

q2 = -26.84 uC ------------<<<<<<<<<<<<<<<<smaller charge