If charge and distance between two charges are reduced to half. Force between them

Question

If charge and distance between two charges are reduced to half. Force between them (A) remains same (B) increases four times (C) reduce four times (D) None of the above

Tardigrade Admin · Accepted Answer

As per Coulomb's law F = k (q1 q2/r2) According to question, F'= k ((q1 / 2)(q2 / 2)/(r / 2)2) =k (q1 q2/r2)=F Thus, force between them will remain same.

Q. If charge and distance between two charges are reduced to half. Force between them

remains same

increases four times

reduce four times

None of the above

As per Coulomb's law $F = k \frac{q _{1} q _{2}}{r ^{2}}$
According to question,
$F^{'} = k \frac{( q _{1} /2 ) ( q _{2} /2 )}{( r /2 ) ^{2}}$
$= k \frac{q _{1} q _{2}}{r ^{2}} = F$
Thus, force between them will remain same.

Q. If charge and distance between two charges are reduced to half. Force between them

remains same

increases four times

reduce four times

None of the above

As per Coulomb's law F=kr2q1​q2​​ According to question, F′=k(r/2)2(q1​/2)(q2​/2)​ =kr2q1​q2​​=F Thus, force between them will remain same.

As per Coulomb's law $F = k \frac{q _{1} q _{2}}{r ^{2}}$
According to question,
$F^{'} = k \frac{( q _{1} /2 ) ( q _{2} /2 )}{( r /2 ) ^{2}}$
$= k \frac{q _{1} q _{2}}{r ^{2}} = F$
Thus, force between them will remain same.