(a) Three charged balls of fluff are nailed down to the table at coordinates (0.30 m, 0),(0, 0.10 m), and (0, 0). They have charges 6.0, -1.0, and 5.0 nanocoulombs. How largeis the force on the charge at the origin due to the other charges? (b) What direction is itin? (c) If you vacuumed up the charge at the origin to get rid of it, is there still an electricfield there? What is its magnitude and direction?

Answered: (a) Three charged balls of fluff are…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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(a) Three charged balls of fluff are nailed down to the table at coordinates (0.30 m, 0),
(0, 0.10 m), and (0, 0). They have charges 6.0, -1.0, and 5.0 nanocoulombs. How large
is the force on the charge at the origin due to the other charges? (b) What direction is it
in? (c) If you vacuumed up the charge at the origin to get rid of it, is there still an electric
field there? What is its magnitude and direction?

Expert Solution

Step 1:Introduction

The electric forces acting on the charge at the origin is shown as below:

The electric force acting on the vertical direction is the attractive force acting in the negative y-direction , which is , $F_{12} = - |\frac{k q_{1} q_{2}}{{r_{12}}^{2}}| \hat{j}$ and the electric force acting on the horizontal direction is the repulsive force acting in the positive x-direction , which is , $F_{23} = |\frac{k q_{2} q_{3}}{{r_{23}}^{2}}| \hat{i}$ , where $k = 9 \times 10^{9} N m^{2} / C^{2}$ is the Coulomb's constant , $q_{1} = - 1 \times 10^{- 9} C$ , $q_{2} = 5 \times 10^{- 9} C$ , $q_{3} = 6 \times 10^{- 9} C$ , $r_{12} = 0.1 m$ and $r_{23} = 0.3 m$ .

The magnitude of the net electric force is calculated by using the formula $F_{2} = \sqrt{{(F_{12})}^{2} + {(F_{23})}^{2}}$ and the direction is calculated by using the formula $θ = \tan^{- 1} (\frac{F_{12}}{F_{23}})$ .

The electric field acting at the origin is shown as below:

The electric field acting on the vertical direction is the field acting in the positive y-direction , which is , $E_{1} = |\frac{k q_{1}}{{r_{12}}^{2}}| \hat{j}$ and the electric field acting on the horizontal direction is the field acting in the negative x-direction , which is , $E_{3} = - |\frac{k q_{3}}{{r_{23}}^{2}}| \hat{i}$ , where $k = 9 \times 10^{9} N m^{2} / C^{2}$ is the Coulomb's constant , $q_{1} = - 1 \times 10^{- 9} C$ , $q_{2} = 5 \times 10^{- 9} C$ , $q_{3} = 6 \times 10^{- 9} C$ , $r_{12} = 0.1 m$ and $r_{23} = 0.3 m$ .

The magnitude of the net electric field is calculated by using the formula $E_{2} = \sqrt{{(E_{1})}^{2} + {(E_{3})}^{2}}$ and the direction is calculated by using the formula $ϕ = \tan^{- 1} (\frac{E_{1}}{E_{3}})$ .

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