Chapter 24, Problem 90P

(a)

To determine

ToCalculate: The potential difference between the cylinders.

Answer to Problem 90P

  $V=\frac{2kQ\ln \left(b/a\right)}{\kappa L}$

Explanation of Solution

Given information :

Radii of coaxial conducting thin cylindrical shells = $a\&b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $-Q$

Formula used :

Electric potential:

  $V=\frac{Q}{C}$

Where, Q is the charge stored and C is the capacitance.

  $C=\frac{2\pi {\in }_0\kappa L}{\ln \left(b/a\right)}$

Where, $\kappa$ is the dielectric constant, ${\in }_0$ is the absolute permittivity and L is the length of the capacitor, a and b are radii of the

Calculation:

The potential difference between the cylinders to their charge and capacitance is,

  $V=\frac{Q}{C}$

The capacitance of a cylindrical capacitor as a function of its radii a and b and length L:

  $C=\frac{2\pi {\in }_0\kappa L}{\ln \left(b/a\right)}$

  $\begin{array}{l}V=\frac{Q\ln \left(b/a\right)}{2\pi {\in }_0\kappa L}\\ V=\frac{2kQ\ln \left(b/a\right)}{\kappa L}\end{array}$

Conclusion:

The potential difference between the cylinders is $V=\frac{2kQ\ln \left(b/a\right)}{\kappa L}$ .

(b)

To determine

ToCalculate: The density of the free charge of on the inner cylinder and the outer cylinder.

Answer to Problem 90P

  ${\sigma }_f\left(a\right)=\frac{Q}{2\pi aL}$

  ${\sigma }_f\left(b\right)=\frac{-Q}{2\pi bL}$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a\&b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $-Q$

Formula used:

Charge density:

  $\sigma =\frac{Q}{2\pi rL}$

Where, Q is the charge, r is the radius and L is the length of the cylinder.

Calculation:

Surface charge density is:

  ${\sigma }_f\left(a\right)=\frac{Q}{2\pi aL}$

  ${\sigma }_f\left(b\right)=\frac{-Q}{2\pi bL}$

Conclusion:

The density of the free charge of on the inner cylinder and the outer cylinder are:

  ${\sigma }_f\left(a\right)=\frac{Q}{2\pi aL}$

  ${\sigma }_f\left(b\right)=\frac{-Q}{2\pi bL}$

(c)

To determine

ToCalculate: The bound charge density ${\sigma }_b$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

Answer to Problem 90P

  ${\sigma }_b\left(a\right)=\frac{-Q\left(\kappa -1\right)}{2\pi aL\kappa }$

  ${\sigma }_b\left(b\right)=\frac{-Q\left(\kappa -1\right)}{2\pi bL\kappa }$

Explanation of Solution

Given information :

Radii of coaxial conducting thin cylindrical shells = $a\&b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $-Q$

Formula used :

Bound charge can be expressed as:

  $Q_b=\frac{Q\left(\kappa -1\right)}{\kappa }$

Where, $\kappa$ is the dielectric constant.

Bound charge density:

  ${\sigma }_b=\frac{Q_b}{A}$

Where, A is the area.

Calculation:

  $Q_b\left(a\right)=\frac{-Q\left(\kappa -1\right)}{\kappa }$

  $Q_b\left(b\right)=\frac{Q\left(\kappa -1\right)}{\kappa }$

  $\begin{array}{l}{\sigma }_b\left(a\right)=\frac{Q_b\left(a\right)}{A}\\ {\sigma }_b\left(a\right)=\frac{\frac{-Q\left(\kappa -1\right)}{\kappa }}{2\pi aL}\\ {\sigma }_b\left(a\right)=\frac{-Q\left(\kappa -1\right)}{2\pi aL\kappa }\end{array}$

  $\begin{array}{l}{\sigma }_b\left(b\right)=\frac{Q_b\left(b\right)}{A}\\ {\sigma }_b\left(b\right)=\frac{\frac{-Q\left(\kappa -1\right)}{\kappa }}{2\pi bL}\\ {\sigma }_b\left(b\right)=\frac{-Q\left(\kappa -1\right)}{2\pi bL\kappa }\end{array}$

Conclusion:

The bound charge density ${\sigma }_b$ on the inner cylindrical surface of the dielectric and on the outer cylindrical surface of the dielectric.

  ${\sigma }_b\left(a\right)=\frac{-Q\left(\kappa -1\right)}{2\pi aL\kappa }$

  ${\sigma }_b\left(b\right)=\frac{-Q\left(\kappa -1\right)}{2\pi bL\kappa }$

(d)

To determine

ToCalculate: The total stored energy.

Answer to Problem 90P

  $U=\frac{k{Q}^2\ln \left(b/a\right)}{\kappa L}$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a\&b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $-Q$

Formula used:

The energy stored in the capacitor:

  $U=\frac{1}{2}QV$

Where, Q is the charge and V is the electric potential.

Calculation:

The potential difference between the cylinders is $V=\frac{2kQ\ln \left(b/a\right)}{\kappa L}$ .

The total stored energy in terms of the charge stored and the potential difference between the cylinders:

  $\begin{array}{l}U=\frac{1}{2}QV\\ U=\frac{1}{2}Q\left[\frac{2kQ\ln \left(b/a\right)}{\kappa L}\right]\\ U=\frac{k{Q}^2\ln \left(b/a\right)}{\kappa L}\end{array}$

Conclusion:

The total stored energy is,

  $U=\frac{k{Q}^2\ln \left(b/a\right)}{\kappa L}$

(e)

To determine

ToCalculate: Mechanical work that is required to remove the dielectric cylindrical shell.

Answer to Problem 90P

  $W=\frac{k{Q}^2\left(\kappa -1\right)\ln \left(b/a\right)}{\kappa L}$

Explanation of Solution

Given information:

Radii of coaxial conducting thin cylindrical shells = $a\&b$

Charge of the inner cylinder = $+Q$

Charge of the outer cylinder = $-Q$

Formula used:

Work done in terms of potential energy of the system:

  $W=\Delta U$

Calculation:

The work required to remove the dielectric cylindrical shell in terms of the change in the potential energy of the system:

  $\begin{array}{l}W=\Delta U\\ W=U^{\prime }-U\end{array}$

The potential energy of the system with the dielectric shell in place is,

  $U^{\prime }=\kappa U$

  $\begin{array}{l}W=\kappa U-U\\ W=U\left(\kappa -1\right)\\ W=\frac{k{Q}^2\left(\kappa -1\right)\ln \left(b/a\right)}{\kappa L}\end{array}$

Conclusion:

Mechanical work that is required to remove the dielectric cylindrical shell is $W=\frac{k{Q}^2\left(\kappa -1\right)\ln \left(b/a\right)}{\kappa L}$ .