Question

Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are doubled and their separation is halved, what happens to the electrical force that each charge exerts on the other one

Answer 1

F' = 16 F

Hence, the electric force between charges becomes sixteen times its initial value.

Step-by-step explanation:

The electric force between the two charges is given by the Colomb's Law:

`F = (KQ_1Q_2)/(R^2)` ------------------- eq(1)

where

F = electric force

K = Colomb's Constant

Q₁ = magnitude of the first charge

Q₂ = magnitude of the second charge

R = Distance between charges

Now the magnitudes of the charges are doubled and the distance between them is halved. Therefore:

`F' = (K(2Q_!)(2Q_2))/(((R)/(2))^2)\\\\F' = 16 (KQ_1Q_2)/(R^2)`

using equation (1):

F' = 16 F

Hence, the electric force between charges becomes sixteen times of its initial value.