Two charged spheres 10 cm apart attract each other with a force of 3.0 x 10^-6 N. What electrostatic force will result if both charges are doubled and the distance remains the same?

The electrostatic force that will result if both charges are doubled and the distance remains the same is 6.0 * 10⁻⁶ N

Explanation:

Coulomb's law of  electricity states that the magnitude of the force of attraction or repulsion between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance of of separation between them.

In formula; F = KQ₁Q₂/d²

Using the formula for electrostatic force of attraction above to determine the force of between the two charged spheres

Let the charges be Q₁ and Q₂; distance of separation be d, K is a constant

Initially, F₁ = KQ₁Q₂/d² (1)

F₁ = 3.0 * 10⁻⁶ N

when the charges are doubled, Q₁ = 2Q₁; Q₂ = 2Q₂; K and d remains constant

F₂ = 2KQ₁Q₂/d² (2)

Dividing equation (2) by (1) to find the ratio of their forces

F₂/F₁ = (2KQ₁Q₂/d²) / KQ₁Q₂/d²

F₂/F₁ = 2

Thus, F₂ is twice F₁.

Since F₁ = 3.0 * 10⁻⁶ N; F₂ = 2 * 3.0 * 10⁻⁶ N

F₂ = 6.0 * 10⁻⁶ N

Therefore, the electrostatic force that will result if both charges are doubled and the distance remains the same is 6.0 * 10⁻⁶ N

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.