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

Explanation:

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 @$1.8\cdot 10^{-7}N@$ .