Suggested languages for you:

Americas

Europe

Q14P

Expert-verifiedFound in: Page 680

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** ${\mathbf{\varphi}}{\mathbf{=}}{\mathbf{5}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{5}}}{{\mathbf{Nm}}}^{{\mathbf{2}}}{\mathbf{/}}{\mathbf{C}}$**. (a) What is the charge of the central particle? What are the net charges of (b) shell A and (c) shell B?**

- The charge of the central particle is $1.77\mathrm{\mu C}$.
- The charge of shell A is $-5.3\mathrm{\mu C}$.
- The charge of shell B is $+8.9\mathrm{\mu C}$.

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

**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)

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

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

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

The next value that takes the flux is given as:

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

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

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

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

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

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

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

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

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

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

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 $+8.9\mathrm{\mu C}$ .

94% of StudySmarter users get better grades.

Sign up for free