Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Flux and non-conducting shells. A charged particle is suspended at the center of two concentric spherical shells that are very thin and made of non-conducting material. Figure 23-37a shows a cross section. Figure 23-37b gives the net flux $ϕ$ through a Gaussian sphere centered on the particle, as a function of the radius r of the sphere. The scale of the vertical axis is set by $ϕ = 5.0 \times 10^{5} {Nm}^{2} / C$ . (a) What is the charge of the central particle? What are the net charges of (b) shell A and (c) shell B?

The charge of the central particle is $1.77 μC$ .
The charge of shell A is $- 5.3 μC$ .
The charge of shell B is $+ 8.9 μC$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

The scale of the vertical axis, $ϕ_{s} = 5.0 \times 10^{5} N . m^{2} / C$

Step 2: Understanding the concept of Gauss law-planar symmetry

Using the concept of the Gauss law, we can get the charge enclosed by the surface. This value of the enclosed charge will determine the charge of the shells of A and B.

Formula:

The enclosed charge by the surface, $q_{enc} = ε_{0} ϕ$ (1)

Step 3: a) Calculation of the charge of the central particle

From the graph, the value $ϕ_{s} = 2.0 \times 10^{5} {Nm}^{2} / C$ for small r leads to the charge of the central particle using equation (1i) as:

$q_{centre} = (8.85 \times 10^{- 12} C^{2} / N . m^{2}) (2.0 \times 10^{5} N . m^{2} / C) = 1.77 \times 10^{- 6} C$

Hence, the value of the charge is $1.77 \times 10^{- 6} C$ .

Step 4: b) Calculation of the charge of shell A

The next value that takes the flux is given as:

$ϕ_{s} = - 4.0 \times 10^{5} N . m^{2} / C$

which implies that the charge enclosed by it using equation (1) is given as:

$q_{enc} = - 3.54 \times 10^{- 6} C$

But we have already accounted for some of that charge in part (a), so the result for part (b) is calculated as follows:

$q_{A} = q_{enc} - q_{centre} = - 5.3 \times 10^{- 6} C$

Hence, the value of the charge is $- 5.3 \times 10^{- 6} C$ .

Step 5: c) Calculation of the charge of shell B

Finally, for the large r-value, the value of the flux is given as:

$ϕ = 6.0 \times 10^{5} N . m^{2} / C$

which implies that the enclosed charge for this flux value is given using equation (1) as:

role="math" localid="1657346976925" $q_{totalence} = 5.31 \times 10^{- 6} C$

Considering what we have already found, then the result of the charge of the shell is given as:

$q_{B} = q_{total enc} - q_{A} = q_{centre} = + 8.9 μC$

Hence, the value of the charge is $+ 8.9 μC$ .

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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 Gauss law-planar symmetry

Step 3: a) Calculation of the charge of the central particle

Step 4: b) Calculation of the charge of shell A

Step 5: c) Calculation of the charge of shell B

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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 3: a) Calculation of the charge of the central particle

Step 4: b) Calculation of the charge of shell A

Step 5: c) Calculation of the charge of shell B

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