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?

F'=  10F (N)

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)