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

(a) The electrostatic force on the 3.78-nC particle is
`10.35 * 10^(-6) N` along the

positive x axis.

(b) The electrostatic force on the 1.95-nC particle is
`10.35 * 10^(-6) N` along the

negative x axis.

Step-by-step explanation:

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

`F_1 = (Q_1 * 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 = (1.95 * 10^(-9) * 3.78 * 10^(-9))/(4 * \pi * 8.854 \time 10^(-12) * (0.08)^2) = 10.35 * 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.