Chapter 6, Problem 6.43P

To determine

The solution of Laplace' equation.

Answer to Problem 6.43P

The potential field $V_1$ is $\frac{-{\rho }_0{r}^2}{2\epsilon }-\frac{{\epsilon }_0{V}_o{c}^3}{r\left[\epsilon \left(b-c\right)-c^2{\epsilon }_0\right]}+V_o\left(\frac{r^3}{2\left(b-c\right)}\right)+\frac{V_o{\epsilon }_0\left(\frac{1}{{\epsilon }_0}-\frac{1}{\epsilon }\right)}{r\left(\frac{\epsilon }{c^2{\epsilon }_0}-\frac{1}{c^2}\right)}$ and $V_2$ is $V_0\frac{\left(b-r\right)}{b-c}-\frac{V_0\epsilon \left(b-r\right)}{\left[\epsilon \left(b-c\right)-c^2{\epsilon }_0\right]}$.

Explanation of Solution

Calculation:

The Poisson's equation (generalization of Laplace equation) is defined for $r<c$ boundary condition.

The Poisson's equation is written as

   $\begin{array}{c}{\nabla }^2{V}_1=\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d{V}_1}{dr}\right)\\ =-\frac{{\rho }_0}{\epsilon }\end{array}$ ...... (1)

Here,

   $\nabla V$ is the potential difference.

   ${\rho }_0$ is the surface charge density.

   ${\epsilon }_0$ is the absolute permittivity.

Simplify equation (1) as,

   $\frac{d}{dr}\left(r^2\frac{d{V}_1}{dr}\right)=-\frac{{\rho }_0{r}^2}{\epsilon }$ ...... (2)

Integrate equation (2) with respect to spherical coordinate $r$ as

   $\frac{d{V}_1}{dr}=-\frac{{\rho }_{0}r}{3\epsilon }+\frac{C_1}{r^2}$ ...... (3)

Substitute $c$ for $z$ in equation (3).

   $\frac{d{V}_1}{dr}=-\frac{{\rho }_{0}c}{3\epsilon }+\frac{C_1}{c^2}$ ...... (4)

Simplified the equation (4) as,

   $\epsilon \frac{d{V}_1}{dr}=-\frac{{\rho }_{0}c}{3}+\frac{\epsilon C_1}{c^2}$ ...... (5)

The Laplace's equation is defined for $r>c$ boundary condition.

The Laplace's equation is written as

   $\frac{d^2{V}_2}{d{r}^2}=0$ ...... (5)

Integrate the equation (5) with respect to spherical coordinate $r$ as

   $\frac{d{V}_2}{dr}=C_1{}^{\prime }$ ..... (6)

Equation (6) is multiplied with ${\epsilon }_0$ on both sides as

   ${\epsilon }_0\frac{d{V}_2}{dr}={\epsilon }_0{C}_1{}^{\prime }$ ...... (7)

If equation (6) and equation (7) is equal, then it is written as

   $-\frac{{\rho }_{0}c}{3}+\frac{\epsilon C_1}{c^2}={\epsilon }_0{C}_1{}^{\prime }$ ..... (8)

Simplified the equation (8) as,

   $C_1{}^{\prime }=\frac{-{\rho }_{0}c}{3{\epsilon }_0}+\frac{\epsilon C_1}{c^2{\epsilon }_0}$ ...... (9)

Substitute $\frac{-{\rho }_{0}c}{3{\epsilon }_0}+\frac{\epsilon C_1}{c^2{\epsilon }_0}$ for $C_1{}^{\prime }$ in equation (6).

   $\frac{d{V}_2}{dr}=\frac{-{\rho }_{0}c}{3{\epsilon }_0}+\frac{\epsilon C_1}{c^2{\epsilon }_0}$ ..... (10)

At boundary condition $r=c$ the derivative of potential field of Poisson's theorem and Laplace's theorem should be equal as,

   $\frac{d{V}_1}{dr}=\frac{d{V}_2}{dr}$ ....... (11)

Substitute $-\frac{{\rho }_{0}c}{3\epsilon }+\frac{C_1}{c^2}$ for $\frac{d{V}_1}{dr}$ and $\frac{-{\rho }_{0}c}{3{\epsilon }_0}+\frac{\epsilon C_1}{c^2{\epsilon }_0}$ for $\frac{d{V}_2}{dr}$ in equation (11)

   $-\frac{{\rho }_{0}c}{3\epsilon }+\frac{C_1}{c^2}=\frac{-{\rho }_{0}c}{3{\epsilon }_0}+\frac{\epsilon C_1}{c^2{\epsilon }_0}$ ...... (12)

Simplify equation (12) as

   $C_1=\frac{\frac{{\rho }_{0}c}{3}\left(\frac{1}{{\epsilon }_0}-\frac{1}{\epsilon }\right)}{\left(\frac{\epsilon }{c^2{\epsilon }_0}-\frac{1}{c^2}\right)}$

Integrate the equation (3) with respect to $z$ as

   $V_1=\frac{-{\rho }_0{r}^2}{6\epsilon }-\frac{C_1}{r}+C_2$ ...... (13)

Substitute $\frac{\frac{{\rho }_{0}c}{3}\left(\frac{1}{{\epsilon }_0}-\frac{1}{\epsilon }\right)}{\left(\frac{\epsilon }{c^2{\epsilon }_0}-\frac{1}{c^2}\right)}$ for $C_1$ and $0$ for $V_1$ in equation (13).

   $0=\frac{-{\rho }_0{r}^2}{6\epsilon }-\frac{\frac{{\rho }_{0}c}{3}\left(\frac{1}{{\epsilon }_0}-\frac{1}{\epsilon }\right)}{r\left(\frac{\epsilon }{c^2{\epsilon }_0}-\frac{1}{c^2}\right)}+C_2$ ...... (14)

Simplify equation (14) as

   $C_2=\frac{{\rho }_0{r}^2}{6\epsilon }+\frac{\frac{{\rho }_{0}c}{3}\left(\frac{1}{{\epsilon }_0}-\frac{1}{\epsilon }\right)}{r\left(\frac{\epsilon }{c^2{\epsilon }_0}-\frac{1}{c^2}\right)}$

Integrate the equation (10) with respect to $z$ as

   $V_2=\frac{-{\rho }_{0}cr}{3{\epsilon }_0}+\frac{\epsilon C_{1}r}{c^2{\epsilon }_0}+C_2{}^{\prime }$ ..... (15)

Substitute $b$ for $r$ and $0$ for $V_2$ in equation (15).

   $0=\frac{-{\rho }_{0}bc}{3{\epsilon }_0}+\frac{\epsilon C_{1}b}{c^2{\epsilon }_0}+C_2{}^{\prime }$ ..... (16)

Equation (16) is simplified as

   $C_2{}^{\prime }=\frac{{\rho }_{0}bc}{3{\epsilon }_0}-\frac{\epsilon C_{1}b}{c^2{\epsilon }_0}$

Substitute $\frac{{\rho }_0{r}^2}{6\epsilon }+\frac{\frac{{\rho }_{0}c}{3}\left(\frac{1}{{\epsilon }_0}-\frac{1}{\epsilon }\right)}{r\left(\frac{\epsilon }{c^2{\epsilon }_0}-\frac{1}{c^2}\right)}$ for $C_2$ in equation (13).

   $V_1=\frac{-{\rho }_0{r}^2}{2\epsilon }-\frac{C_1}{r}+\frac{{\rho }_0{r}^2}{6\epsilon }+\frac{\frac{{\rho }_{0}c}{3}\left(\frac{1}{{\epsilon }_0}-\frac{1}{\epsilon }\right)}{r\left(\frac{\epsilon }{c^2{\epsilon }_0}-\frac{1}{c^2}\right)}$ ...... (17)

Substitute $\frac{{\rho }_{0}bc}{3{\epsilon }_0}-\frac{\epsilon C_{1}b}{c^2{\epsilon }_0}$ for $C_2{}^{\prime }$ in equation (15).

   $\begin{array}{c}{V}_2=\frac{-{\rho }_{0}cr}{3{\epsilon }_0}+\frac{\epsilon C_{1}r}{c^2{\epsilon }_0}+\frac{{\rho }_{0}bc}{3{\epsilon }_0}-\frac{\epsilon C_{1}b}{c^2{\epsilon }_0}\\ =\frac{{\rho }_{0}c}{3{\epsilon }_0}\left(b-r\right)-\frac{\epsilon C_1}{c^2{\epsilon }_0}\left(b-r\right)\end{array}$ ...... (18)

If equation (17) and equation (18) is equal, then it is written as

   $\frac{-{\rho }_0{r}^2}{2\epsilon }-\frac{C_1}{r}+\frac{{\rho }_0{r}^2}{6\epsilon }+\frac{\frac{{\rho }_{0}c}{3}\left(\frac{1}{{\epsilon }_0}-\frac{1}{\epsilon }\right)}{r\left(\frac{\epsilon }{c^2{\epsilon }_0}-\frac{1}{c^2}\right)}=\frac{{\rho }_{0}c}{3{\epsilon }_0}\left(b-r\right)-\frac{\epsilon C_1}{c^2{\epsilon }_0}\left(b-r\right)$ ..... (19)

Substitute $c$ for $r$ in equation (19).

   $C_1=\frac{{\rho }_0{c}^4\left(b-c\right)}{3\left[\epsilon \left(b-c\right)-c^2{\epsilon }_0\right]}$

Substitute $\frac{{\rho }_0{c}^4\left(b-c\right)}{3\left[\epsilon \left(b-c\right)-c^2{\epsilon }_0\right]}$ for $C_1$ in equation (17)

   $V_1=\frac{-{\rho }_0{r}^2}{2\epsilon }-\frac{{\rho }_0{c}^4\left(b-c\right)}{3r\left[\epsilon \left(b-c\right)-c^2{\epsilon }_0\right]}+\frac{{\rho }_0{r}^2}{6\epsilon }+\frac{\frac{{\rho }_{0}c}{3}\left(\frac{1}{{\epsilon }_0}-\frac{1}{\epsilon }\right)}{r\left(\frac{\epsilon }{c^2{\epsilon }_0}-\frac{1}{c^2}\right)}$ ...... (20)

Substitute $\frac{{\rho }_0{c}^4\left(b-c\right)}{3\left[\epsilon \left(b-c\right)-c^2{\epsilon }_0\right]}$ for $C_1$ in equation (18)

   $V_2=\frac{{\rho }_{0}c}{3{\epsilon }_0}\left(b-r\right)-\frac{{\rho }_0\epsilon c^2\left(b-c\right)\left(b-r\right)}{3{\epsilon }_0\left[\epsilon \left(b-c\right)-c^2{\epsilon }_0\right]}$ ...... (21)

The value of potential difference $V_0$ is,

   $V_0=\frac{{\rho }_{0}c}{3{\epsilon }_0}\left(b-c\right)$

Substitute $\frac{{\rho }_{0}c}{3{\epsilon }_0}\left(b-c\right)$ for $V_0$ in equation (20).

   $V_1=\frac{-{\rho }_0{r}^2}{2\epsilon }-\frac{{\epsilon }_0{V}_o{c}^3}{r\left[\epsilon \left(b-c\right)-c^2{\epsilon }_0\right]}+V_o\left(\frac{r^3}{2\left(b-c\right)}\right)+\frac{V_o{\epsilon }_0\left(\frac{1}{{\epsilon }_0}-\frac{1}{\epsilon }\right)}{r\left(\frac{\epsilon }{c^2{\epsilon }_0}-\frac{1}{c^2}\right)}$

Substitute $\frac{{\rho }_{0}c}{3{\epsilon }_0}\left(b-c\right)$ for $V_0$ in equation (21).

   $V_2=V_0\frac{\left(b-r\right)}{b-c}-\frac{V_0\epsilon \left(b-r\right)}{\left[\epsilon \left(b-c\right)-c^2{\epsilon }_0\right]}$

Conclusion:

Therefore, the potential field $V_1$ is $\frac{-{\rho }_0{r}^2}{2\epsilon }-\frac{{\epsilon }_0{V}_o{c}^3}{r\left[\epsilon \left(b-c\right)-c^2{\epsilon }_0\right]}+V_o\left(\frac{r^3}{2\left(b-c\right)}\right)+\frac{V_o{\epsilon }_0\left(\frac{1}{{\epsilon }_0}-\frac{1}{\epsilon }\right)}{r\left(\frac{\epsilon }{c^2{\epsilon }_0}-\frac{1}{c^2}\right)}$ and the potential field $V_2$ is $V_0\frac{\left(b-r\right)}{b-c}-\frac{V_0\epsilon \left(b-r\right)}{\left[\epsilon \left(b-c\right)-c^2{\epsilon }_0\right]}$.