Question

Two point charges q1 and q2 are 3.00 m apart, and their total charge is 20 μC . (a) If the force of repulsion between them is 0.075N, what are magnitudes of the two charges? (b) If one charge attracts the other with a force of 0.525N, what are the magnitudes of the two charges? Note that you may need to solve a quadratic equation to reach your answer.

Answer 1

Explanation:

Let two charges are $q_1\ and\ q_2$, such that, $q_1+q_2=20\ \mu C$.............(1)

Distance between charges, r = 3 m

The force of repulsion between them is, F = 0.075 N

We can consider that both charges are positive, so that there exists the force of repulsion. The force of repulsion is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$q_1q_2=\dfrac{Fr^2}{k}$

$q_1q_2=\dfrac{0.075\times 3^2}{9\times 10^9}$

$q_1q_2=7.5\times 10^{-11}$.............(2)

On solving equation (1) and (2) we get :

$q_1=0.382\ \mu C$

$q_2=19.618\ \mu C$

(b) Now the force of attractive, F = -0.525 N

Let charges are $q_1\ and\ -q_2$ such that, $q_1-q_2=20\ \mu C$.............(3)

Force of attraction is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$q_1q_2=\dfrac{Fr^2}{k}$

$q_1q_2=\dfrac{-0.525\times 3^2}{9\times 10^9}$

$q_1q_2=-5.25\times 10^{-10}$.............(4)

Now on solving equation (3) and (4) we get :

$q_1=0.266\ \mu C$

and

$q_2=-19.734\ \mu C$

Hence, this is the required solution.