Particle A of charge 2.97 x 10^-4 C is at the origin, particle B of charge -5.94 x 10^-4 C is at...

Question:

Particle {eq}A {/eq} of charge {eq}2.97 \times 10^{-4} \, \mathrm{C} {/eq} is at the origin, particle {eq}B {/eq} of charge {eq}-5.94 \times 10^{-4} \, \mathrm{C} {/eq} is at {eq}(4.00 \, \mathrm{m}, 0) {/eq}, and particle {eq}C {/eq} of charge {eq}1.04 \times 10^{-4} \, \mathrm{C} {/eq} is at {eq}(0, 3.00 \, \mathrm{m}) {/eq}. We wish to find the net electric force on {eq}C {/eq}.

(a) What is the {eq}x {/eq}-component of the electric force exerted by {eq}A {/eq} on {eq}C {/eq}?

(b) What is the {eq}y {/eq}-component of the force exerted by {eq}A {/eq} on {eq}C {/eq}?

(c) Find the magnitude of the force exerted by {eq}B {/eq} on {eq}C {/eq}.

(d) Calculate the {eq}x {/eq}-component of the force exerted by {eq}B {/eq} on {eq}C {/eq}.

(e) Calculate the {eq}y {/eq}-component of the force exerted by {eq}B {/eq} on {eq}C {/eq}.

(f) Sum the two {eq}x {/eq}-components from parts (a) and (d) to obtain the resultant {eq}x {/eq}-component of the electric force acting on {eq}C {/eq}.

(g) Similarly, find the {eq}y {/eq}-component of the resultant force acting on {eq}C {/eq}.

(h) Find the magnitude and direction of the resultant electric force acting on {eq}C {/eq}.

Electric Force:

From Coulomb's law, we get that the magnitude of the electric force exerted by a charge Q on another point charge q a distance r away is given by

$$F \ = \ k \ \frac{|Q| \ |q|}{r^2}$$

where k is the Coulomb constant.

Answer and Explanation:

The magnitude of the electric force exerted by a charge Q on another point charge q a distance r away is

$$F \ = \ k \ \frac{|Q| \ |q|}{r^2}$$

whe...

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Electric Force: Definition & Equation

from High School Physical Science: Tutoring Solution

Chapter 14 / Lesson 12
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