Question

Two equal charges placed in air are separated by a distance 3 m repel each other with force 0.1gf. Calculate the magnitude of eitger of the charges

Answer 1

The force between two particles is given by the equation

F=k×(q1)(q2)/r*2

Know knowing k's value 9×10*9 we calculate the charge and 1gf=0.0098N

=0.001

Answer 2

Answer:

Concept:

Coulomb’s Law (F=kqq/r2) is an equation that represents the attractive or repulsive electric force (F) of two-point charges (q). The force is proportional to the square of the distance (r) between charges.  The proportionality constant that relates force to charge and distance is given the letter k, or ke, and is known as Coulomb’s constant.

Explanation:

Given:

          $d=3 m, f=0.19 f$

By using coulombs law,      $\mathrm{f}=\frac{\mathrm{kq}^{2}}{\mathrm{r}^{2}}$

        $\mathrm{kq}^{2}=0.19 \times 9$

        $\mathrm{q}^{2}=\frac{1.71}{9.0 \times 10^{9}}$

where, ${k}=9.0 \times 10^{9}$

          $\mathrm{q}=\sqrt{0.19 \times 10^{-9}} \mathrm{C}$

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