Question

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?

Answer 1

Answer:

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

Answer 2

Electrostatic force is a force imposed by one charge on another as a result of the field.Electrostatic force will  be 12.0 x 10⁻⁶ N if both charges are doubled and the distance remains the same.

What is the electrostatic force?

It is a force imposed by one charge on another as a result of the field.

The electrostatic force produced by one line charge on another line charge separated by distance d is determined by the charge potency of each charge as well as the separation distance between them.

Hence the electrostatic force is given by  

$\rm F=\frac{Kq_1q_2}{d^2}$

The given data in the question ,

d is the distance between the charge

F₁ is the electric force for case 1= 3.0 x 10⁻⁶ N

F₂ is the  electric force for case 2= ?

Conditions for case 2;

(q₁=2q₁),(q₂=2q₂),d₂=d₁

For case 1,

$\rm F_1=\frac{Kq_1q_2}{d_1^2}$

For case 2,

$\rm F_2=\frac{K(2q_1)(2q_2)}{d_1^2}$

$\rm F_2=4\times\frac{Kq_1q_2}{d_1^2}\\\\\rm F_2=4F_1$

$\rm F_2=4\times\frac{Kq_1q_2}{d_1^2}\\\\\rm F_2=4\times3.0\times 10^{-6}\\\\\rm F_2=12.0\times10^{-6}$

Hence electrostatic force will be 12.0 x 10⁻⁶ N.  if both charges are doubled and the distance remains the same.

To learn more about the electrostatic force refer to the link;

https://brainly.com/question/9774180