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

The magnitude of the charges can be found using Coulomb's Law and setting up equations based on the given information. For the force of repulsion of 0.075N, the charges have magnitudes of approximately 17.967 μC and 2.033 μC. For the force of attraction of 0.525N, the charges have magnitudes of approximately 8.150 μC and 11.850 μC.

Step-by-step explanation:

(a) To find the magnitude of the charges, we can use Coulomb's Law:

F = k * (|q1| * |q2|) / r^2

where F is the force of repulsion, k is the electrostatic constant, |q1| and |q2| are the magnitudes of the charges, and r is the distance between them.

Plugging in the given values, we have:

0.075 = (8.99 * 10^9) * (|q1| * |q2|) / (3.00^2)

Simplifying and solving the equation, we find that |q1| * |q2| = 2.4995 * 10^-9 C^2

Since the charges have a total charge of 20 μC, we can set up the equation |q1| + |q2| = 20 * 10^-6 C.

Solving these two equations simultaneously, we find that the magnitudes of the charges are approximately |q1| = 17.967 μC and |q2| = 2.033 μC.

(b) Using the same equation as in part (a), we can set up the equation 0.525 = (8.99 * 10^9) * (|q1| * |q2|) / (3.00^2), and solve for |q1| and |q2| using a quadratic equation.

The magnitudes of the charges in this case are approximately |q1| = 8.150 μC and |q2| = 11.850 μC.

Answer 2

Step-by-step 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(q_1q_2)/(r^2)`

`q_1q_2=(Fr^2)/(k)`

`q_1q_2=(0.075* 3^2)/(9* 10^9)`

`q_1q_2=7.5* 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(q_1q_2)/(r^2)`

`q_1q_2=(Fr^2)/(k)`

`q_1q_2=(-0.525* 3^2)/(9* 10^9)`

`q_1q_2=-5.25* 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.