Question

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.

Answer 1

Answer:

Option (a) is correct.

Explanation:

Let the magnitude of the positive charge = $q_1$ and the magnitude of the negative charge = $q_2$

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

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

$F=k\frac {q_1 q_2}{d^2}$, where k is a constant.

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

$F_i = k\frac {q_1 q_2}{1^2}= k q_1q_2\cdots(i)$

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, $q_1$ as well as $q_2$, are remain the same.

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

$F_f = k\frac {q_1 q_2}{1^2}= k q_1q_2$

From equation (i), $F_f=F_i$

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.

Answer 2

When the electrostatic force lies between the charges so here the force should remain attractive, also the magnitude should remain the same.

Impact on the electrostatic force:

Here we assume the magnitude of the positive charge should be q1 and the magnitude of the negative charge should be q2

Also, the distance between the charge should be d = 1 m

So here we use the columb law

F = kq1q2/d2

here k should be constant

Here the nature of the force should be attractive since the both the charges should be opposite.

Hence, the first option is correct.

Learn more about force here: https://brainly.com/question/3398162