Question

Two point charges are separated by a distance
1 m experiences 100 N of electrostatic force. If
the separation between them becomes 0.25 m
then force between them will be​

Answer 1

Answer:

force is inversely proportional to the square of distance between charges so when distance is reduced by 4 times force increased by 16 times 1600N

Answer 2

Answer:

The force between two same charges separated by a distance  of 1 m is 100N, If  the separation between them becomes 0.25 m  then the force between them will be​ 1557.5N

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

let the charge be $q$

the distance between the charges $r=1m$

the amount of force $F=100N$

put  the given values in the equation for force

$100=\frac{9*10^{9}*q*q }{1^{2} }$

⇒

$100=9*10^{9}*q^{2} \\q^{2}=0.11*10^{-7}\\q=1.04*10^{-4} C$

new  distance $r=0.25m$

new force $F=\frac{9*10^{9}*1.04*10^{-4}*1.04*10^{-4} }{(0.25)^{2} }$

⇒$F=1557.5N$