Question

Two point charges q1 and q2 are 3m apart and their combined charges are 20 micron meter . if one repel the other with a force of 0.075 newton what are the two charges

Answer 1

The values of the two charges are  5.01 x 10^-6 C  and  14.99 x 10^-6 C.

Explanation:

q1 +q2 = 20 μC

Distance between charges = 3 m

Force of repulsion between charges =  0.075 N

• Now according to coulomb's law

F = Kq1q2/ r^2

0.075 =  (8.99 x 10^9 x) q1q2 / 3.0

q1q2 = (0.075) (3)/  (8.99 x 10^9 x)

q1q2 = 7.51 x 10^-11 C^2 ----(2)

• Converting  μC to C

q1 = 20 μC ( 1C/ 10^6C) - q2 ----(1)

q1 = 20 x 10^6 C

q1q2 = 7.51 x 10^-11 C^2 ----(2)

q1 = 20 μC ( 1C/ 10^6C) - q2 ----(1)

(20 x 10^6 - q2) q2 = 7.51 x 10^-11 C^2

(20 x 10^6)q2 - q2^2 = 7.51 x 10^-11 C^2

By solving quadratic equation

q2 = 14.99 x 10^-6 C or 5.01 x 10^-6 C

q1 =  5.01 x 10^-6 C or 14.99 x 10^-6 C

Thus the values of the two charges are  5.01 x 10^-6 C  and  14.99 x 10^-6 C.