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Expert-verified Found in: Page 680 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # 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 ${\mathbf{\varphi }}$ 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 ${\mathbf{\varphi }}{\mathbf{=}}{\mathbf{5}}{\mathbf{.}}{\mathbf{0}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{5}}}{{\mathbf{Nm}}}^{{\mathbf{2}}}{\mathbf{/}}{\mathbf{C}}$. (a) What is the charge of the central particle? What are the net charges of (b) shell A and (c) shell B? 1. The charge of the central particle is $1.77\mathrm{\mu C}$.
2. The charge of shell A is $-5.3\mathrm{\mu C}$.
3. The charge of shell B is $+8.9\mathrm{\mu C}$.
See the step by step solution

## Step 1: The given data

The scale of the vertical axis, ${\mathrm{\varphi }}_{\mathrm{s}}=5.0×{10}^{5}\mathrm{N}.{\mathrm{m}}^{2}/\mathrm{C}$

## 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, ${\mathrm{q}}_{\mathrm{enc}}={\mathrm{\epsilon }}_{0}\mathrm{\varphi }$ (1)

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

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

${\mathrm{q}}_{\mathrm{centre}}=\left(8.85×{10}^{-12}{\mathrm{C}}^{2}/\mathrm{N}.{\mathrm{m}}^{2}\right)\left(2.0×{10}^{5}\mathrm{N}.{\mathrm{m}}^{2}/\mathrm{C}\right)\phantom{\rule{0ex}{0ex}}=1.77×{10}^{-6}\mathrm{C}$

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

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

The next value that takes the flux is given as:

${\mathrm{\varphi }}_{\mathrm{s}}=-4.0×{10}^{5}\mathrm{N}.{\mathrm{m}}^{2}/\mathrm{C}$

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

${\mathrm{q}}_{\mathrm{enc}}=-3.54×{10}^{-6}\mathrm{C}$

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

${\mathrm{q}}_{\mathrm{A}}={\mathrm{q}}_{\mathrm{enc}}-{\mathrm{q}}_{\mathrm{centre}}\phantom{\rule{0ex}{0ex}}=-5.3×{10}^{-6}\mathrm{C}$

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

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

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

$\mathrm{\varphi }=6.0×{10}^{5}\mathrm{N}.{\mathrm{m}}^{2}/\mathrm{C}$

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

role="math" localid="1657346976925" ${\mathrm{q}}_{\mathrm{totalence}}=5.31×{10}^{-6}\mathrm{C}$

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

${\mathrm{q}}_{\mathrm{B}}={\mathrm{q}}_{\mathrm{total}\mathrm{enc}}-{\mathrm{q}}_{\mathrm{A}}={\mathrm{q}}_{\mathrm{centre}}\phantom{\rule{0ex}{0ex}}=+8.9\mathrm{\mu C}\phantom{\rule{0ex}{0ex}}$

Hence, the value of the charge is $+8.9\mathrm{\mu C}$ . ### Want to see more solutions like these? 