Question

If two equal charges each IC were seperated in air by a distance of 1km what would be the force between them​

Answer 1

Answer:

9000 N

Explanation:

From the formula,

$kq_{1} q_{2} /r^{2}$

So as both charges are 1 C each and separated by 1km = 1000m,

The repulsive force between them is $9*10^9(1)(1)/(1000*1000)\\= 9*10^9*(10^-6) = 9000N$

Therefore, there a repulsive force (because both charges are positive in nature and like charges repel) of 9000 N between them.