Question

$+2\text{C}$ and $+6\text{C}$ charges are repelling each other with a force of $12\text{N}$. If each charge is given $-2\text{C}$ of charge, then the value of the force will be:-

• A. $4\text{N}$ (attractive)
• B. $4\text{N}$ (repulsive)
• C. $8\text{N}$
• D. zero

Answer 1

The correct option is D zero

If each charge is given $-2\text{C}$ of charge then, the magnitude of new charges is,

$q_1=2C+(-2C)=0$
$q_2=6C+(-2C)=4C$

So, the now the force between the new charges is,

$F=\frac{K{q}_1{q}_2}{r^2}=\frac{K\times 0\times 4}{r^2}$

$F=0\text{N}$