Question

Two charge bodies, Q1 and Q2 having charges ‘X’ coulombs and ‘Y’ coulombs respectively, are far apart at the distance of 28meters. They are facing some imbalance in their amount of positive and negative charges which is 0.05%. Estimate the electrostatic force of attraction or repulsion between them. Where X = 58 & Y= 41​

a) Attraction force of 0.064 N

b) Repulsion force of 0.064 N

c) Attraction force of 0.128 N

d) Repulsion force of 0.128 N

Answer 1

The electrostatic force of attraction or repulsion between two charged bodies can be calculated using Coulomb's Law.

So, the correct answer is (a) Attraction force of 0.064 N.

Step-by-step explanation:

The electrostatic force of attraction or repulsion between two charged bodies can be calculated using Coulomb's Law, which states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In this case, let's calculate the electrostatic force between the charged bodies Q1 and Q2. Given that X = 58C and Y = 41C, the force can be calculated using the formula:

F = (k * (Q1 * Q2)) / r^2

Plugging in the values, we get:

F = (9 * 10^9) * (58 * 41) / (28^2)

Simplifying this, we find that the force is approximately 0.064 N.

So, the correct answer is (a) Attraction force of 0.064 N.