$(A)\;5:4\\ (B)\;4:5 \\ (C)\; 1:5 \\ (D)\;5:1 $

$F_{\large before }=\large\frac{1}{4 \pi \in _0} \frac{(5) \times (-25)}{r^2}$

Charge after $=q_1=q_2=\large\frac{5+ (-25)}{2}$

$\qquad=-10$

$F_{\large after }=\large\frac{1}{4 \pi \in _0} \frac{(-10) \times (-10)}{2}$

$\large\frac{|F_{\Large after}|}{|F_{\Large before}|}=\bigg| \large\frac{(-10) \times (-10)}{(5) \times (-25)}\bigg|$

$\qquad= \large\frac{4}{5}$

$\qquad=4:5$

Hence B is the correct answer.

Ask Question

Tag:MathPhyChemBioOther

Take Test

...