When the distance between two charged particles is halved. The coulomb force between them becomes :

Question

When the distance between two charged particles is halved. The coulomb force between them becomes : (A) four times (B) double (C) one half (D) one fourth

Tardigrade Admin · Accepted Answer

Earlier force between two particles is F =( kq 1 q 2/ r 2) When distance is halved force becomes F '=( kq 1 q 2/( r / 2)2)=4 ( kq 1 q 2/ r 2)=4 F

Q. When the distance between two charged particles is halved. The coulomb force between them becomes :

four times

double

one half

one fourth

Earlier force between two particles is
$F = \frac{k q _{1} q _{2}}{r ^{2}}$
When distance is halved force becomes
$F^{'} = \frac{k q _{1} q _{2}}{( r /2 ) ^{2}} = 4 \frac{k q _{1} q _{2}}{r ^{2}} = 4 F$

Q. When the distance between two charged particles is halved. The coulomb force between them becomes :

four times

double

one half

one fourth

Earlier force between two particles is F=r2kq1​q2​​ When distance is halved force becomes F′=(r/2)2kq1​q2​​=4r2kq1​q2​​=4F

Earlier force between two particles is
$F = \frac{k q _{1} q _{2}}{r ^{2}}$
When distance is halved force becomes
$F^{'} = \frac{k q _{1} q _{2}}{( r /2 ) ^{2}} = 4 \frac{k q _{1} q _{2}}{r ^{2}} = 4 F$