Question

An attractive force of $9$ N acts between $+5$ μC and $-5$ μC at some distance. These charges are allowed to touch each other and are then again placed at their initial position. The force acting between them will be :

• A. infinite
• B. $9\times {10}^9$ N
• C. $1$ N
• D. zero

Answer 1

Answer: (D)

zero

According to the question

$F=\frac{1}{4\pi {ϵ}_0}\frac{(5\times {10}^{-6})\times (-5\times {10}^{-6})}{r^2}$

On touching them the charge on each will be zero.

$\therefore$ $F=\frac{1}{4\pi {ϵ}_0}\frac{\left(0\right)}{r^2}=0$