Question

1. What is the magnitude of the force on a charge of +40 μC that is 0.6 m from a charge of - 80 μC?

Answer 1

Answer:

The magnitude of the force is 79.893 N.

Explanation:

The magnitude of the electrostatic force between the two particles is determined by Coloumb's Law, whose formula is:

$F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}$ (1)

Where:

$\kappa$ - Electrostatic constant, in newtons-square meters per square coulomb.

$q_{A}, q_{B}$ - Electric charges, in coulombs.

$r$ - Distance, in meters.

If we know that $\kappa = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$, $q_{A} = +40\times 10^{-6}\, C$, $q_{B} = - 80\times 10^{-6}\, C$ and $r = 0.6\,m$, then the magnitude of the force is:

$F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}$

$F = 79.893\,N$

The magnitude of the force is 79.893 N.