A positive charge is 1 m to the left of a negative charge. The positive charge is then moved and placed 1 m to the right of the negative charge. What happens to the electrostatic force between the charges? The electrostatic force remains attractive, and the magnitude does not change.
The electrostatic force remains repulsive, and the magnitude does not change.
The electrostatic force remains repulsive, but the magnitude changes.
The electrostatic force remains attractive, but the magnitude changes.

Option (a) is correct.

Explanation:

Let the magnitude of the positive charge = and the magnitude of the negative charge =

Earlier, the distance between both the charge, d = 1m

By using Coulomb's law, the magnitude of the force between two charge

, where k is a constant.

So, the magnitude of the force in the initial configuration,

The nature of the force is attractive, as both the charges are opposite in nature.

On replacing the positive charge from 1m left to 1 m right side of the negative charge, the distance between the charges remains the same, i.e d=1m.

Moreover, the magnitude, and nature of each charge, as well as , are remain the same.

So, the magnitude of the force in the final configuration,

From equation (i),

The nature of the force is attractive, as both the charges are opposite in nature.

So, the electrostatic force remains attractive, and the magnitude does not change.

Hence, option (a) is correct.

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

read more on -