Question

Two identical point charges exert a repulsive force of 0.500 N on one another when separated by 1.5 m. What is the magnitude of the net charge of either point charge?

Answer 1

Given,

The repulsive force exerted by the charges, F=0.500 N

The distance between the charges, d=1.5 m

From Coulomb's law,

`F=\frac{\text{kqq}}{r^2}`

Where q is the magnitude of the charge of each point charge and k is the coulomb's constant.

On rearranging the above equation,

`\begin{gathered} F=(kq^2)/(r^2) \\ \Rightarrow q=\sqrt[]{(F)/(k)}r \end{gathered}`

On substituting the known values,

`\begin{gathered} q=\sqrt[]{(0.5)/(9*10^9)}*1.5 \\ =1.1*10^(-5)\text{ C} \end{gathered}`

Thus the magnitude of the charge of each point charge is 1.1×10⁻⁵ C

Therefore the correct answer is option B.