Question

The electric field due to a charged particle at a point 1.0 m away from it has
magnitude 81NC-1.What is the magnitude of the electric charge on the particle?
What is the magnitude of the electrostatic force on a particle having the same
charge kept at a distance of 2.0 m from it?

Answer 1

The electric field due to a charged particle according to Coulomb's law is $E=k\frac {q}{r^2}$ where $k=9\cdot10^9Nm^2C^{-2}$. Thus $q=\frac{E\cdot r^2}{k}=\frac{81 NC^{-1}\cdot1m^2}{9\cdot 10^9 Nm^2C^{-2}}=9\cdot 10^{-9}C$ is chage of the particle. The magnitude of the electrostatic force between two particles with chage $q$ at $R=2.0m$ is

$F=k\frac{q^2}{R^2}=9\cdot 10^9\frac{81\cdot 10^{-18}}{4}=1.8\cdot 10^{-7}N$

Answer: The magnitude of the electric charge on the particle is $9\cdot 10^{-9}C$ . The magnitude of the electrostatic force on a particle having the same charge kept at a distance of 2.0 m from it is $1.8\cdot 10^{-7}N$ .