Question

A 1.95-nC charged particle located at the origin is separated by a distance of 0.0800 m from a 3.78-nC charged particle located farther along the positive x axis. Both particles are held at their locations by an external agent.
(a) What is the electrostatic force on the 3.78-nC particle?
(b) What is the electrostatic force on the 1.95-nC particle?

Answer 1

Answer:

(a) The electrostatic force on the 3.78-nC particle is $10.35 \times 10^{-6} N$ along the  

    positive x axis.

(b) The electrostatic force on the 1.95-nC particle is $10.35 \times 10^{-6} N$ along the  

    negative x axis.

Explanation:

a.) The electrostatic force on the 3.78 nC particle is

    $F_1 = \frac{Q_1 \times Q_2}{4 \pi \epsilon_0 r^2} \hat{i}$    where $Q_1$ is the 1.95 nC charge at the origin, and

                                        $Q_2$ is the 3.78 nC  charge at 0.08 m from the origin

                                         r is the distance of 0.08 between the charges

                                         $\epsilon_{0}$ is the relative permittivity of 8.854 × $10^{-12} F/m$.

Therefore $F_1 = \frac{1.95 \times 10^{-9} \times 3.78 \times 10^{-9}}{4 \times \pi \times 8.854 \time 10^{-12} \times (0.08)^2} = 10.35 \times 10^{-6} Newton$ in the positive x direction.

b.) The electrostatic force on the 1.95 nC particle is the same as the electrostatic force on the 3.78 nC particle except that its direction is in the negative x- axis direction.