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.

Explanation:

Let two charges are , such that, .............(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 :

.............(2)

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

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

Let charges are such that, .............(3)

Force of attraction is given by :

.............(4)

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

and

Hence, this is the required solution.

chemical energy --> gravitational potential energy --> kinetic energy --> elastic potential energy --> gravitational potential energy

explanation:

when emma starts the hike, the chemical energy stored in her body is used to climb up the mountain, and so she gains gravitational potential energy (because at the end, she will be at a higher height h above the ground). then she goes bungee jumping: as she falls, this gravitational potential energy is converted into kinetic energy (because her speed increases as she falls down), until the point when the rope stretches, and she stops moving. at that moment, all the kinetic energy is converted into elastic potential energy due to the stretching of the rope. then, this energy is released because the rope returns to its natural length, and emma gains a small increase in altitude , so she gains some gravitational potential energy.