Question

A charge of uniform linear density 3.00 nC/m is distributed along a long, thin, nonconducting rod. The rod is coaxial with a long conducting cylindrical shell (inner radius = 4.40 cm, outer radius = 10.6 cm). The net charge on the shell is zero. (a) What is the magnitude (in N/C) of the electric field at distance r = 15.0 cm from the axis of the shell? What is the surface charge density on the (b) inner and (c) outer surface of the shell?

Answer 1

Concept and Reason

The required concept to solve this is electric field due to the infinitely long cylinder, linear charge density, and surface charge density.

Use the Gauss’ law to solve for the expression of electric field at any distance outside the cylindrical conductor.

Use the relationship between the linear charge density and the surface area of the shell to find the surface charge density on the inner surface of the shell.

The surface charge density on the outer surface of the shell is calculated by using the relationship between linear charge density and the area of the rod.

Fundamentals

The total electric flux through the Gaussian surface is equal to the total charge enclosed by the surface S.

The magnitude of the electric field is calculated as follows:

$\int {E \cdot ds = \left( {\frac{Q}{{{\varepsilon _0}}}} \right)}$

Here, E is the magnitude of electric field, ds is the surface area, Q is the charge enclosed by the surface, and ${\varepsilon _0}$ is the permittivity of free space.

The magnitude of the surface charge density of the inner surface of the shell is,

${\sigma _{{\rm{in}}}} = \left( {\frac{{ - \lambda }}{{2\pi r}}} \right)$

Here, $\lambda$ is the linear surface charge density and r is the radius of the shell.

The magnitude of the surface charge density of the outer surface of the shell is,

${\sigma _{{\rm{in}}}} = \left( {\frac{\lambda }{{2\pi r}}} \right)$

Here, $\lambda$ is the linear surface charge density and r is the radius of the shell.

(a)

Consider the formula for calculating the electric flux through the spherical surface.

The expression for calculating the electric flux through a closed surface is calculated as follows:

$\phi = \int {E \cdot ds}$

Here, E is the magnitude of electric field and ds is the surface area of the spherical surface.

Summing over the whole thin nonconducting rod, the flux through the linear charge rod is calculated as follows:

$\begin{array}{l}\\\phi = \int {E \cdot ds} \\\\ = E\int {ds} \\\\ = E \times 2\pi rl\\\end{array}$

Here, l is the length of the rod.

Find the magnitude of charge enclosed by the Gaussian surface.

The expression for calculating the charge enclosed by the surface by using the Gauss’s law as follows:

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

Here, Q is the charge enclosed by the surface and ${\varepsilon _0}$ is the permittivity of free space.

Substitute $\left( {\lambda l} \right)$ for Q in equation as follows:

$\begin{array}{l}\\\phi = \frac{Q}{{{\varepsilon _0}}}\\\\ = \left( {\frac{{\lambda l}}{{{\varepsilon _0}}}} \right)\\\end{array}$

Here, $\lambda$ is the linear charge density and $\left( l \right)$ is the length of the rod.

Find the magnitude of electric field by using Gauss’s law.

The expression for calculating the magnitude of electric field by using Gauss’s law is as follows:

$\begin{array}{c}\\\phi = E \times 2\pi rl\\\\ = \left( {\frac{{\lambda l}}{{{\varepsilon _0}}}} \right)\\\\E = \left( {\frac{\lambda }{{2\pi r{\varepsilon _0}}}} \right)\\\end{array}$

Substitute $\left( {3.00 \times {{10}^{ - 9}}{\rm{ C/m}}} \right)$ for $\lambda$ , $\left( {8.85 \times {{10}^{ - 12}}{\rm{ }}{{\rm{C}}^2}/{\rm{N}} \cdot {{\rm{m}}^2}} \right)$ for ${\varepsilon _0}$ , and $15.0{\rm{ cm = }}\left( {\frac{{\left( {15.0{\rm{ m}}} \right)\left( {1.00{\rm{ m}}} \right)}}{{100{\rm{ cm}}}}} \right)$ for r in equation as follows:

$\begin{array}{c}\\E = \left( {\frac{\lambda }{{2\pi r{\varepsilon _0}}}} \right)\\\\ = \left( {\frac{{3.00 \times {{10}^{ - 9}}{\rm{ C/m}}}}{{2\left( {3.14} \right)\left( {15.0 \times {{10}^{ - 2}}{\rm{ m}}} \right)\left( {8.85 \times {{10}^{ - 12}}{\rm{ }}{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}} \right)}}} \right)\\\\ = 0.359 \times {10^3}{\rm{ N/C}}\\\end{array}$

(b)

Calculate the surface charge density on the inner surface of the shell.

The formula for calculating the surface charge density of the inner surface of the shell as follows:

${\sigma _{{\rm{in}}}} = \left( {\frac{{ - \lambda }}{{2\pi r}}} \right)$

Here, $\lambda$ is the linear surface charge density and r is the radius of the shell.

Substitute $\left( {3.00 \times {{10}^{ - 9}}{\rm{ C/m}}} \right)$ for $\lambda$ and $\left( {0.044{\rm{ m}}} \right)$ for r in equation as follows:

$\begin{array}{c}\\{\sigma _{{\rm{in}}}} = \left( {\frac{{ - \lambda }}{{2\pi r}}} \right)\\\\ = \left( {\frac{{ - 3.00 \times {{10}^{ - 9}}{\rm{ C/m}}}}{{2\left( {3.14} \right)\left( {0.044{\rm{ m}}} \right)}}} \right)\\\\ = - \left( {10.85 \times {{10}^{ - 9}}{\rm{ C}}} \right)\\\end{array}$

(c)

Calculate the surface charge density on the outer surface of the shell.

The formula for calculating the surface charge density of the inner surface of the shell as follows:

$\begin{array}{c}\\{\sigma _{{\rm{out}}}} = \left( {\frac{q}{{2\pi rl}}} \right)\\\\ = \frac{{\lambda l}}{{2\pi rl}}\\\\ = \left( {\frac{\lambda }{{2\pi r}}} \right)\\\end{array}$

Here, $\lambda$ is the linear surface charge density and r is the radius of the shell.

Substitute $\left( {3.00 \times {{10}^{ - 9}}{\rm{ C/m}}} \right)$ for $\lambda$ and $\left( {0.106{\rm{ m}}} \right)$ for r in equation as follows:

$\begin{array}{c}\\{\sigma _{{\rm{in}}}} = \left( {\frac{\lambda }{{2\pi r}}} \right)\\\\ = \left( {\frac{{3.00 \times {{10}^{ - 9}}{\rm{ C/m}}}}{{2\left( {3.14} \right)\left( {0.106{\rm{ m}}} \right)}}} \right)\\\\ = 4.504 \times {10^{ - 9}}{\rm{ C}}\\\end{array}$

Ans: Part a

The magnitude of the electric field at distance $r = 15.0{\rm{ cm}}$ from the axis of the shell is $0.359 \times {10^3}{\rm{ N/C}}$ .

Part b

The magnitude of the surface charge density on the inner surface of the shell is $\left( { - 10.85{\rm{ }} \times {\rm{1}}{{\rm{0}}^{ - {\rm{9}}}}{\rm{ C/}}{{\rm{m}}^{\rm{2}}}} \right)$ .

Part c

The magnitude of the surface charge density on the outer surface of the shell is $4.504 \times {10^{ - 9}}{\rm{ C}}$ .