Chapter 24, Problem 86P

(a)

To determine

ToCalculate: The electric field in the regions $a<r<\frac{1}{2}\left(a+b\right)$ and $\frac{1}{2}\left(a+b\right)<r<b$ .

Answer to Problem 86P

  $\begin{array}{l}{E}_1=\frac{kQ}{{\kappa }_1{r}^2}\\ E_2=\frac{kQ}{{\kappa }_2{r}^2}\end{array}$

Explanation of Solution

Given information :

The radius of spherical capacitor $=a$

Charge $=+Q$

Inner radius of conducting spherical shell $=b$

Charge $=-Q$

Dielectric constants $={\kappa }_1\&{\kappa }_2$

Formula used :

Electric field:

  $E=\frac{kQ}{\kappa r^2}$

Where, k is the Coulomb’s constant, Q is the charge, $\kappa$ is the dielectric constant and r is the distance between the plates.

Calculation:

The electric fields in the regions is $a<r<\frac{1}{2}\left(a+b\right)$ and $\frac{1}{2}\left(a+b\right)<r<b$ is:

  $\begin{array}{l}{E}_1=\frac{kQ}{{\kappa }_1{r}^2}\\ E_2=\frac{kQ}{{\kappa }_2{r}^2}\end{array}$

Conclusion:

The electric field in the regions $a<r<\frac{1}{2}\left(a+b\right)$ and $\frac{1}{2}\left(a+b\right)<r<b$ is,

  $\begin{array}{l}{E}_1=\frac{kQ}{{\kappa }_1{r}^2}\\ E_2=\frac{kQ}{{\kappa }_2{r}^2}\end{array}$

(b)

To determine

ToCalculate: TheIntegrate expression of $dV=-\overrightarrow{E}\cdot d\overrightarrow{l}$ .

Answer to Problem 86P

  $\left|V\right|=\frac{kQ\left(b-a\right)}{\left(a+b\right)}\frac{{\kappa }_{1}a+{\kappa }_{2}b}{{\kappa }_1{\kappa }_{2}ab}$

Explanation of Solution

Given information:

The radius of spherical capacitor $=a$

Charge $=+Q$

Inner radius of conducting spherical shell $=b$

Charge $=-Q$

Dielectric constants $={\kappa }_1\&{\kappa }_2$

Formula used:

  $\left|V\right|=\left|V_1-V_2\right|$

Calculation:

Because the electric field is directed radially away from the center, the potential of the outer spherical shell is less than the potential of the inner sphere. Hence, the difference in potential between the spheres is given by:

  $\begin{array}{l}\left|V\right|=\left|V_1-V_2\right|\\ \left|V\right|=\left|-{\displaystyle \underset{a}{\overset{\frac{1}{2}\left(a+b\right)}{\int }}{E}_{1}dr-\left(-{\displaystyle \underset{\frac{1}{2}\left(a+b\right)}{\overset{b}{\int }}{E}_{2}dr}\right)}\right|\\ \left|V\right|=\left|-\frac{kQ}{{\kappa }_1}{\displaystyle \underset{a}{\overset{\frac{1}{2}\left(a+b\right)}{\int }}{r}^{-2}dr+\frac{kQ}{{\kappa }_2}{\displaystyle \underset{\frac{1}{2}\left(a+b\right)}{\overset{b}{\int }}{r}^{-2}dr}}\right|\end{array}$

  $\begin{array}{l}\left|V\right|={\left|{\left[\frac{kQ}{{\kappa }_1}\frac{1}{r}\right]}_a^{\frac{1}{2}\left(a+b\right)}+\frac{kQ}{{\kappa }_2}\frac{1}{r}\right|}_{\frac{1}{2}\left(a+b\right)}^b\\ \left|V\right|=\left|\frac{kQ}{{\kappa }_1}\left(\frac{2}{a+b}-\frac{1}{a}\right)+\frac{kQ}{{\kappa }_2}\left(\frac{1}{b}-\frac{2}{a+b}\right)\right|\\ \left|V\right|=\left|\frac{kQ\left(a-b\right)}{\left(a+b\right)}\frac{{\kappa }_{1}a+{\kappa }_{2}b}{{\kappa }_1{\kappa }_{2}ab}\right|\\ \left|V\right|=\frac{kQ\left(b-a\right)}{\left(a+b\right)}\frac{{\kappa }_{1}a+{\kappa }_{2}b}{{\kappa }_1{\kappa }_{2}ab}\end{array}$

Conclusion:

TheIntegrate expression of $dV=-\overrightarrow{E}\cdot d\overrightarrow{l}$ is $\left|V\right|=\frac{kQ\left(b-a\right)}{\left(a+b\right)}\frac{{\kappa }_{1}a+{\kappa }_{2}b}{{\kappa }_1{\kappa }_{2}ab}$ .

(c)

To determine

ToCalculate: An expression for the capacitance of the system.

Answer to Problem 86P

  $C=\frac{Q}{\frac{kQ\left(a-b\right)}{a+b}\frac{{\kappa }_{1}b+{\kappa }_{2}b}{{\kappa }_1{\kappa }_{2}ab}}$

Explanation of Solution

Given information :

The radius of spherical capacitor $=a$

Charge $=+Q$

Inner radius of conducting spherical shell $=b$

Charge $=-Q$

Dielectric constants $={\kappa }_1\&{\kappa }_2$

Formula used :

The capacitance of the capacitor:

  $C=\frac{Q}{\left|V\right|}$

Q is the charge stored and V is the voltage.

Calculation:

  $\begin{array}{l}C=\frac{Q}{\left|V\right|}\\ C=\frac{Q}{\frac{kQ\left(a-b\right)}{a+b}\frac{{\kappa }_{1}b+{\kappa }_{2}b}{{\kappa }_1{\kappa }_{2}ab}}\end{array}$

Conclusion:

An expression for the capacitance of the system is,

  $C=\frac{Q}{\frac{kQ\left(a-b\right)}{a+b}\frac{{\kappa }_{1}b+{\kappa }_{2}b}{{\kappa }_1{\kappa }_{2}ab}}$

(d)

To determine

ToShow:The answer from Part (c) simplifies to the expected one if the ${\kappa }_1$ equals ${\kappa }_2$ .

Explanation of Solution

Given information:

The radius of spherical capacitor $=a$

Charge $=+Q$

Inner radius of conducting spherical shell $=b$

Charge $=-Q$

Dielectric constants $={\kappa }_1\&{\kappa }_2$

Formula used:

The charge stored in the capacitor:

  $Q=CV$

Where, C is the capacitance and V is the voltage.

Calculation:

  $\left|V\right|=\frac{kQ\left(b-a\right)}{\left(a+b\right)}\frac{{\kappa }_{1}a+{\kappa }_{2}b}{{\kappa }_1{\kappa }_{2}ab}$

  $\begin{array}{l}Q=CV\\ C=\frac{Q}{V}\end{array}$

  ${\kappa }_1={\kappa }_2=\kappa$

  $\begin{array}{l}C=\frac{\kappa \kappa ab\left(a+b\right)}{k\left(b-a\right)\left(\kappa a+\kappa b\right)}\\ C=\frac{\kappa ab}{k\left(b-a\right)}\\ C=\frac{4\pi {\in }_0\kappa ab}{b-a}\end{array}$