Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

In Fig. 21-26, particle 1 of charge $+ 1.0 μ C$ and particle 2 of charge $- 3.0 μ C$ are held at separation L=10.0cm on an x-axis. If particle 3 of unknown charge q₃ is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the (a) x and (b) y coordinates of particle 3?

The x-coordinate of particle 3 for the net force on it to be zero is 14cm
The y-coordinate of particle 3 for the net force on it to be zero is 0cm

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

The values of charges on the three particles are:

$q_{1} = + 1.0 μ C, q_{2} = - 3.0 μ C, a n d q_{3} = u n k n o w n$

The separation between particles 1 and 2, $L = 10.0 c m$

The net force on the particle from particles 1 and 2 is zero.

Step 2: Understanding the concept of Coulomb’s law

Using the concept of Coulomb's law, we can get the required equation to solve to get the coordinates of particle 3 for the condition of the net force acting on it as zero.

Formula:

The magnitude of the electrostatic force between any two particles,

$F = k \frac{| q_{1} | | q_{2} |}{r^{2}}$ (1)

Step 3: a) Calculation of the x-coordinate of the location of particle 3

There is no equilibrium position for q₃ between the two fixed charges because it is being pulled by one and pushed by the other (since q₁and q₂ have different signs); in this region, this means the two force arrows on q₃are in the same direction and cannot cancel. It should also be clear that off-axis (with the axis defined as that which passes through the two fixed charges) there are no equilibrium positions. On the semi-infinite region of the axis that is nearest q₂and furthest from q₁. an equilibrium position for q₃cannot be found because |q₁| $<$ |q₂| and the magnitude of the force exerted by q₂is everywhere (in that region) stronger than that exerted by q₁on q₃. Thus, we must look in the semi-infinite region of the axis which is nearest q₁and furthest from q₂, where the net force on q₃has magnitude using equation (1) as follows:

$F_{3} = F_{13} + F_{23}$

$= |k \frac{|q_{1} q_{3}|}{L_{0}^{2}} - k \frac{|q_{2} q_{3}|}{{(L + L_{0})}^{2}}| . . . . . . . . . . . . . . . . . . . . . (2)$

with L = 10 cm and L₀ is assumed to be positive. We set this equal to zero, as required by the problem. Now, we obtain the equation (2) by canceling the equal terms as given: $\frac{| q_{1} |}{(L_{0}^{2})} - \frac{| q_{2} |}{(L + L_{0})^{2}} = 0 {(\frac{L + L_{0}}{L_{0}})}^{2} = | \frac{q_{2}}{q_{1}} | {(\frac{L + L_{0}}{L 0})}^{2} = | (- 3.0 μ C) / (- 1.0 μ C) | {(\frac{L + L_{0}}{L_{0}})}^{} = 3.0$

Therefore, we get,

localid="1662643008359" $(L + L_{0}) / L_{0} = \sqrt 3 L_{0} = \frac{L}{(\sqrt 3 - 1)} L_{0} = \frac{10 c m}{(\sqrt 3 - 1)}$

L₀=14cm

Hence, the value of the x-coordinate of particle 3 is 14cm

Step 4: b) Calculation of the y-coordinate of the location of particle 3

As stated above in the calculations of part (a), y = 0

Hence, the value of the y-coordinate of particle 3 is 0cm.

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of Coulomb’s law

Step 3: a) Calculation of the x-coordinate of the location of particle 3

Step 4: b) Calculation of the y-coordinate of the location of particle 3

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Fundamentals Of Physics

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Step by Step Solution

Step 1: The given data

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Step 4: b) Calculation of the y-coordinate of the location of particle 3

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