Question

 If two charges are fixed such that the distance r between them does not change, but the charge q₁ is decreased by half, what is the new force?

two charges are not changed but they are moved twice farther than the original distance r, what becomes of the new electric force?

 

Answer 1

$F_o=\frac{kq_1q_2}{r^2} : r^2=\frac{kq_1q_2}{F_o}$

If q₁ is reduced by half then new force = $\frac{F_o}{2}$

$F_{new}=\frac{k0.5q_1q_2}{r^2}: r^2 =\frac{k0.5q_1q_2}{F_o}$

$\implies \frac{kq_1q_2}{F_o}=\frac{k0.5q_1q_2}{F_{new}}$

$\implies F_o=2F_{new}$

2)$F_o=\frac{kq_1q_2}{r^2}$

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

$\implies F_{new}(2r)^2=F_o r^2$

$\therefore 4F_{new}=F_o$