Question

Two 1 Coulomb charges are kept at 1m distance in air medium. Force of attraction or repulsion between them will be? *​

Answer 1

Given:-

magnitude of two charges=1 C

distance between two charges, d=1 m

To Find:-

Force between two charges

Explanation:-

According to Coulomb's law, force between two charges separated by a distance 'd' is given by:-

$F=k_{e}\times\dfrac{q_{1}\times q_{2}}{d^{2}}$ where,

$k_{e}$ = coulomb's constant=$9\times10^{9}N-m^{2}.C^{-2}$

$q_{1}=q_{2}=1C$

Put all the values in above equation to obtain the value of force.

$\Rightarrow F=9\times10^{9}\times\dfrac{1\times1}{1^{2}}=9\times10^{9}N$