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,

$F_{initial} = \frac{kq_1q_2}{r^2}$

Here,

k = Coulomb's constant

$q_{1,2}$ = Charge at each object

r = Distance between them

As the distance is doubled so,

$F_{final} = \frac{kq_1q_2}{( 2r )^2}$

$F_{final} = \frac{ kq_1q_2}{ 4r^2}$

$F_{final} = \frac{1}{4} \frac{ kq_1q_2}{r^2}$

$F_{final} = \frac{1}{4} F_{initial}$

$\frac{F_{final}}{ F_{initial}} = \frac{1}{4}$

Therefore the factor is 1/4