Chapter 24, Problem 44P

(a)

To determine

ToFind: Expressions for the electric field and energy density as a function of the distance $R$ from the axis.

Answer to Problem 44P

  $E_{R>R_2}=0$

  $u_{R>R_2}=0$

Explanation of Solution

Given information :

Radius of the wire = $R_1$

Length of the wire = $L$

Charge = $+Q$

Inner radius of cylindrical shell = $R_2$

Length of the cylindrical shell = $L$

Charge = $-Q$

Formula used :

From Gauss’s law:

  $E\left(2\pi RL\right)=\frac{Q_{inside}}{{\in }_0}$

  $E=\frac{\lambda }{2\pi {\in }_{0}R}$

Where, E electric field, $Q_{inside}$ is the enclosed charge, R is the radius, L is the length, $\lambda$ is the charge density and ${\in }_0$ is the absolute permittivity.

Energy density is given by:

  $u=\frac{1}{2}{\in }_0{E}_{}^2$

Calculation:

Using Gauss’s law, find the electric field for cylindrical surfaces of radii $R<R_1,R_1<R<R_2$ , and $R>R_2$

  

For $R<R_1$ :

  $E_{R<R_1}\left(2\pi RL\right)=\frac{Q_{inside}}{{\in }_0}=0$

  $E_{R<R_1}=0$

  $E=0$ for $R<R_1$

  $u_{R<R_1}=0$

For $R_1<R<R_2$ :

  $E_{R_1<R<R_2}\left(2\pi RL\right)=\frac{Q_{inside}}{{\in }_0}=\frac{\lambda L}{{\in }_0}$

  $\lambda =$ The linear charge density

  $E_{R_1<R<R_2}=\frac{\lambda }{2\pi {\in }_{0}R}=\frac{2kQ}{RL}$

The energy density in the region $R_1<R<R_2$ is

  $u_{R_1<R<R_2}=\frac{1}{2}{\in }_0{E}_{R_1<R<R_2}^2$

  $\begin{array}{l}{u}_{R_1<R<R_2}=\frac{1}{2}{\in }_0{\left(\frac{2k\lambda }{R}\right)}^2\\ u_{R_1<R<R_2}=\frac{1}{2}{\in }_0{\left(\frac{2kQ}{RL}\right)}^2\\ u_{R_1<R<R_2}=\frac{2k^2{\in }_0{Q}^2}{R^2{L}^2}\end{array}$

For $R>R_2$ :

  $E_{R>R_2}\left(2\pi RL\right)=\frac{Q_{inside}}{{\in }_0}=0$

  $E_{R>R_2}=0$

Because $E=0$ for $R>R_2$

  $u_{R>R_2}=0$

Conclusion:

Electric field

  $E_{R>R_2}=0$

Energy density

  $u_{R>R_2}=0$

(b)

To determine

ToCalculate: The amount ofenergy that resides in a region between the conductors that has a radius $R$ , a thickness $dR$ , and a volume $2\pi rLdR$ .

Answer to Problem 44P

  $dU=\frac{k{Q}^2}{RL}dR$

Explanation of Solution

Given information:

Radius of the wire = $R_1$

Length of the wire = $L$

Charge = $+Q$

Inner radius of cylindrical shell = $R_2$

Length of the cylindrical shell = $L$

Charge = $-Q$

Formula used:

The energy residing in a cylindrical shell between the conductors of radius $R$ , thickness $dR$ , and volume $2\pi RLdR$ :

  $dU=2\pi RLu\left(R\right)dR$

Calculation:

  $\begin{array}{l}dU=2\pi RLu\left(R\right)dR\\ dU=2\pi RL\left(\frac{2k^2{\in }_0{Q}^2}{R^2{L}^2}\right)dr\\ dU=\frac{k{Q}^2}{RL}dR\end{array}$

Conclusion:

The amount ofenergy is:

  $dU=\frac{k{Q}^2}{RL}dR$

(c)

To determine

To Calculate: The total energy stored in the capacitor.

Answer to Problem 44P

  $U=\frac{k{Q}^2}{L}\ln \left(\frac{R_2}{R_1}\right)$

Explanation of Solution

Given information:

Radius of the wire = $R_1$

Length of the wire = $L$

Charge = $+Q$

Inner radius of cylindrical shell = $R_2$

Length of the cylindrical shell = $L$

Charge = $-Q$

Formula used:

Energy stored in the capacitor:

  $U=\frac{1}{2}C{V}^2$

Where, C is the capacitance and V is the potential difference.

Calculation:

Integrate $dU$ from $R=R_1\text{ to }R=R_2$ to obtain:

  $U=\frac{k{Q}^2}{L}{\displaystyle \underset{R_1}{\overset{R_2}{\int }}\frac{dR}{R}=\frac{k{Q}^2}{L}\ln \left(\frac{R_2}{R_1}\right)}$

  $\begin{array}{l}U=\frac{1}{2}C{V}^2\\ U=\frac{1}{2}\frac{Q^2}{C}\\ U=\frac{Q^2}{2\left(\frac{2\pi {\in }_{0}L}{\ln \left(\frac{R_2}{R_1}\right)}\right)}\\ U=\frac{k{Q}^2}{L}\ln \left(\frac{R_2}{R_1}\right)\end{array}$

Conclusion:

The total energy stored in the capacitor is:

  $U=\frac{k{Q}^2}{L}\ln \left(\frac{R_2}{R_1}\right)$