Question

Two charges q1 and q2 are separated by a distance d and exert a force F on each other. What is the new force F', if charge 1 is increased to q'1=5q1, charge 2 decreased to q'2=q2/2, and the distance is decreased to d'=d/2?

Answer 1

The force between two charges can be calculated using Coulomb's Law. We can use the given values to calculate the new force F' and determine its value.

Step-by-step explanation:

The force between two charges is given by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Using the given information, we can calculate the new force F' using the formula:

F' = (k * q'1 * q'2) / (d'^2)

Substituting the given values into the formula, we get:

F' = (k * 5q1 * (q2/2)) / ((d/2)^2)

After solving the equation, we can determine the value of the new force F'.

Answer 2

F'= 10F (N)

Step-by-step explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

F = K*q₁*q₂ / d² Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)

d: distance between the charges in meters(m)

Calculating of the new force F'

Data:

q'₁ = 5q₁

q'₂ = q₂/2

d' = d/2

We apply the Coulomb's law:

F' = K*q'₁*q'₂ / d'²

F'= K*(5q₁)*(q₂/2) / (d/2)²

F'= K*(5q₁*q₂/2) / (d²/4)

F'= K*20q₁*q₂) / (2d²)

F'= 10(K*q₁*q₂) / (d²)

F'= 10(K*q₁*q₂) / (d²) , F = K*q₁*q₂ / d²

F'= 10F (N)