Question

An electron and a proton are separated by a distance of 1.0 m. What happens to the force between them if the electron moves 0.5 m away from the proton?

Answer 1

The electrostatic force between two charges of 1 C each, separated by a distance of 0.5 m can be calculated using Coulomb's Law. The force is inversely proportional to the square of the distance, so if the distance is increased, the force decreases. In this case, if the electron moves 0.5 m away from the proton, the new force can be calculated using Coulomb's Law.

Step-by-step explanation:

The electrostatic force between two charges of 1 C each, separated by a distance of 0.5 m can be calculated using Coulomb's Law. Coulomb's Law states that the electrostatic force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

So, if the distance between the charges is increased to 1 m, the force between them will decrease. The force is inversely proportional to the square of the distance, which means that if the distance is doubled, the force will decrease by a factor of four.

In this case, if the electron moves 0.5 m away from the proton, the new distance becomes 1.5 m. Using Coulomb's Law, we can calculate the new force:

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

Where F is the force, k is the Coulomb's constant (approximately 9 x 10^9 Nm^2/C^2), q1 and q2 are the charges (in this case, both 1 C), and r is the distance (1.5 m). Plugging in these values, we get:

F = (9 x 10^9 Nm^2/C^2) * (1 C * 1 C) / (1.5 m)^2

F ≈ 4 x 10^9 N

Answer 2

Step-by-step explanation:

It is given that, an electron and a proton are separated by a distance of 1.0 m i.e d = 1 m . At this position, F is the force between them

`F=k(q_1q_2)/(1^2)`...............(1)

We need to find effect on force between them if the electron moves 0.5 m away from the proton. Let the force is F'.

`F'=k(q_1q_2)/(0.5^2)`...............(2)

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

`(F)/(F')=(k(q_1q_2)/(1^2))/(k(q_1q_2)/(0.5^2))`

`(F)/(F')=0.25`

F' = 4 F

So, the distance between the electron and the proton is 0.5 m, the new force will be 4 times of previous force.