Question

The force of repulsion between two positive charge of magnitude 1C each separated by a distance of 1m in vacuum is?

Answer 1

Answer:

If two like point charges are separated by 1m and the repulsion force between them is 9. 0x109N, each charge is called 1 Coulomb, shown as 1C

Explanation:

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Answer 2

Answer:

The force of repulsion between two positive charges of magnitude 1C each separated by a distance of 1m in a vacuum is $9*10^{9}N$

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

the amount of charges are $q_{1} =1C, q_{2} =1C$

the distance between them is $r=1m$

therefore the electrostatic force between them is

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