Question

Two particles with charge Q1 and Q2 are seperated by distance d and they exert force F on each other. If the distance between them is reduced to d/3 then the force they will exert on each other is

Answer 1

AnswEr :

From the Question,

• Initial Distance of Separation : d

• Changed Distance of Separation : d/3

• Charges : (Q_1,Q_2)

To finD

New force acting on the charges.

From Cuolomb's Inverse Square Law,

$\sf F = \dfrac{KQ_1 Q_2}{d^2}-----------(1)$

According to the Question,

$\sf F' = \dfrac{KQ_1Q_2}{(\frac{d}{3})^2}-----------(2)$

Dividing equations (1) and (2),

$\dashrightarrow \sf \dfrac{F}{F'} = \dfrac{9d^2}{d^2} \times \cancel{\dfrac{KQ_1Q_2}{KQ_1Q_2}} \\ \\ \dashrightarrow \sf \dfrac{F}{F'} = 9 \\ \\ \dashrightarrow \boxed{\boxed{\sf F' = \dfrac{F}{9}}}$