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jekas [21]
3 years ago
15

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?
Physics
1 answer:
lesya692 [45]3 years ago
3 0

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.

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