Question

A charged particle is held at the center of two concentric conducting spherical shells. Figure 23-35a shows a cross section. Figure 23-35b 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 Φs = 4.2 × 105 N•m2/C. What is the net charge (in Coulombs) of shell B?

Answer 1

Concepts and reason

The main concept used to solve the problem are flux and Gauss’s Law.

Initially, Use the graph to find the net flux through the gaussian sphere centered on the particle. Later, Use the gauss law to calculate the charge of the central particle and calculate the charge on shell A. Finally, use gauss law to calculate the charge in shell B.

Fundamentals

The gauss’s law is expressed as follows:

$\phi = \frac{q}{{{\varepsilon _0}}}$

Here, q is the charge, $\phi$ is the flux, and ${\varepsilon _0}$ is the permittivity of free space.

From the graph,

The net flux through the gaussian sphere centered on the particle is ,

$\phi = - 7.56 \times {10^5}\,{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/C}}$

The charge of the central particle is obtained for small r as follows:

${q_{{\rm{central}}}} = \phi {\varepsilon _0}$

Substitute $- 7.56 \times {10^5}\,{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/C}}$ for $\phi$ and $8.85 \times 1{{\rm{0}}^{ - {\rm{12}}}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}}{{\rm{m}}^{\rm{2}}}$ for ${\varepsilon _0}$ in expression ${q_{{\rm{central}}}} = \phi {\varepsilon _0}$ .

$\begin{array}{c}\\{q_{{\rm{central}}}} = \left( { - 7.56 \times {{10}^5}\,{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\left( {8.85 \times 1{{\rm{0}}^{ - {\rm{12}}}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}}{{\rm{m}}^{\rm{2}}}} \right)\\\\ = - 6.69 \times {10^{ - 6}}\,{\rm{C}}\\\end{array}$

Refer to the graph the net value of flux is expressed as follows:

$\phi = + 3.36 \times {10^5}\,{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/C}}$

The charge of the central particle is obtained for small r as follows:

${q_{{\rm{enc}}}} = \phi {\varepsilon _0}$

Substitute $+ 3.36 \times {10^5}\,{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/C}}$ for $\phi$ and $8.85 \times 1{{\rm{0}}^{ - {\rm{12}}}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}}{{\rm{m}}^{\rm{2}}}$ for ${\varepsilon _0}$ in expression ${q_{{\rm{enc}}}} = \phi {\varepsilon _0}$ .

$\begin{array}{c}\\{q_{{\rm{enc}}}} = \left( { + 3.36 \times {{10}^5}\,{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/C}}} \right)\left( {8.85 \times 1{{\rm{0}}^{ - {\rm{12}}}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}}{{\rm{m}}^{\rm{2}}}} \right)\\\\ = 2.97 \times {10^{ - 6}}\,{\rm{C}}\\\end{array}$

So, the charge on shell A is calculated as follows:

${q_A} = {q_{{\rm{enc}}}} - {q_{{\rm{central}}}}$

Substitute $2.97 \times {10^{ - 6}}\,{\rm{C}}$ ${q_{{\rm{enc}}}}$ and $- 6.69 \times {10^{ - 6}}\,{\rm{C}}$ for ${q_{{\rm{central}}}}$ in expression ${q_A} = {q_{{\rm{enc}}}} - {q_{{\rm{central}}}}$ .

$\begin{array}{c}\\{q_A} = 2.97 \times {10^{ - 6}}\,{\rm{C}} - \left( { - 6.69 \times {{10}^{ - 6}}\,{\rm{C}}} \right)\\\\ = 9.66 \times {10^{ - 6}}\,{\rm{C}}\\\end{array}$

Consider the larger radius shell , the value for flux is expressed as follows:

$\phi = - 1.68 \times {10^5}\,{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/C}}$

The charge of the outer particle is obtained for small r as follows:

${q_{{\rm{total}}}} = \phi {\varepsilon _0}$

Substitute $- 1.68 \times {10^5}\,{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/C}}$ for $\phi$ and $8.85 \times 1{{\rm{0}}^{ - {\rm{12}}}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}}{{\rm{m}}^{\rm{2}}}$ for ${\varepsilon _0}$ in expression ${q_{{\rm{total}}}} = \phi {\varepsilon _0}$ .

$\begin{array}{c}\\{q_{{\rm{total}}}} = \left( { - 1.68 \times {{10}^5}\,{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/C}}} \right)\left( {8.85 \times 1{{\rm{0}}^{ - {\rm{12}}}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}}{{\rm{m}}^{\rm{2}}}} \right)\\\\ = - 1.48 \times {10^{ - 6}}\,{\rm{C}}\\\end{array}$

Thus, the charge on the shell B is calculated as follows:

${q_B} = {q_{{\rm{total}}}} - {q_A} - {q_{{\rm{central}}}}$

Substitute $9.66 \times {10^{ - 6}}\,{\rm{C}}$ for ${q_A}$ , $- 6.69 \times {10^{ - 6}}\,{\rm{C}}$ for ${q_{{\rm{central}}}}$ , and $- 1.48 \times {10^{ - 6}}\,{\rm{C}}$ for ${q_{{\rm{total}}}}$ in expression ${q_B} = {q_{{\rm{total}}}} - {q_A} - {q_{{\rm{central}}}}$ .

$\begin{array}{c}\\{q_B} = - 1.48 \times {10^{ - 6}}\,{\rm{C}} - 9.66 \times {10^{ - 6}}\,{\rm{C}} - \left( { - 6.69 \times {{10}^{ - 6}}\,{\rm{C}}} \right)\\\\ = - 4.45 \times {10^{ - 6}}\,{\rm{C}}\\\end{array}$

Ans:

The net charge (in Coulombs) of shell B is $- 4.45 \times {10^{ - 6}}\,{\rm{C}}$ .