Question

Two equal charges are placed at a distance of 3 m. Calculate and mention what type of
force experiencing between the charges of 40 C.

Answer 1

Answer:

Two equal charges 40C are placed at a distance of 3 m, the force between them is $F=16*10^{11}N$

Explanation:

• The force acting between stationary charges is called electrostatic force

• The coulomb's law gives the magnitude of the force as

        $F=\frac{Kq_{1} q_{2} }{r^{2} }$

        where  $K=9*10^{9}$

        $q_{1} , q_{2}$ - the charges

        $r$- the distance between the charges

From the question, we have given that

the first charge $q_{1}=40C$

the second charge $q_{2} =40C$

the distance between the charges $r=3m$

⇒ the amount of force

$F=\frac{9*10^{9}*40*40 }{3^{2} }$

⇒ $F=16*10^{11}N$