Question

If the distance between two charges is doubled, by what factor is the magnitude of the electric force changed? F_e final/F_e, initial =____

Answer 1

To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,

`F_(initial) = (kq_1q_2)/(r^2)`

Here,

k = Coulomb's constant

`q_(1,2)` = Charge at each object

r = Distance between them

As the distance is doubled so,

`F_(final) = (kq_1q_2)/(( 2r )^2)`

`F_(final) = ( kq_1q_2)/( 4r^2)`

`F_(final) = (1)/(4) ( kq_1q_2)/(r^2)`

`F_(final) = (1)/(4) F_(initial)`

`(F_(final))/( F_(initial)) = (1)/(4)`

Therefore the factor is 1/4