Question

The electrostatic force of repulsion between two equal charges separated by a distance of 2m is 1N. Find the magnitude of each charge?
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Answer 1

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Two equal charges placed in air and separated by a distance of 2 m repel each other with a force of 10^-4kgf . Calculate the magnitude of either of the charges.

Two equal charges placed in air and separated by a distance of 2 m repel each other with a force of 10^-4kgf . Calculate the magnitude of either of the charges.The electrostatic force of repulsion between two equal positively charged ions is 3.7 × 10^-9N , when they are separated by a distance of 5 ∘A . ...

Answer 2

Answer:

The electrostatic force of repulsion between two equal charges separated by a distance of 2m is 1N, the magnitude of each charge is $2.09*10^{-5}C$

​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 $r=2m$

the amount of force $F=1N$

put  the given values in the equation for force

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

⇒

$4=9*10^{9}*q^{2} \\q^{2}=0.44*10^{-9}\\q=2.09*10^{-5} C$