- #1
cwill53
- 220
- 40
- Homework Statement
- This problem is a rehash of Problem 2.50 of the book Introduction to Electrodynamics by Griffiths.
A volume charge density of some configuration is
$$\rho (r)=\epsilon _0A\left [ 4\pi \delta ^3(\mathbf{r})-\frac{\lambda ^2}{r}e^{-\lambda r} \right ]$$
where ##A## and ##\lambda ## are positive constants, and ##r## is the distance from origin in spherical coordinates.
(a) Find the total charge of this configuration.
(b) Calculate the potential due to this ##\rho (r)## by using the expression for the potential
due to a continuous charge distribution.
(c) Calculate the electric field ##E## directly from the potential and verify that the
differential form of Gauss’s law is obeyed.
- Relevant Equations
- $$Q=\int_{\mathbb{R}^3}^{}\rho \, d^3\mathbf{r}$$
$$V(\mathbf{r})=\frac{1}{4\pi \epsilon _0}\int_{\Gamma }^{}\frac{\rho (\mathbf{r'})}{\left \| \mathbf{r-r'} \right \|}d^3\mathbf{r'}$$
I know for sure that the desired potential is
$$V(\mathbf{r})=A\frac{e^{-\lambda r}}{r}$$
Part (a) was simple, after applying
$$Q=\int_{\mathbb{R}^3}^{}\rho \, d^3\mathbf{r}$$
I found that the total charge of the configuration was zero.
Part (b) is where the difficulties arise for me. I applied
$$V(\mathbf{r})=\frac{1}{4\pi \epsilon _0}\int_{\Gamma }^{}\frac{\rho (\mathbf{r'})}{\left \| \mathbf{r-r'} \right \|}d^3\mathbf{r'}$$
to the given charge distribution. Here's my work.
$$V(\mathbf{r})=\frac{1}{4\pi \epsilon_0}\epsilon _0A\left [ 4\pi \int_{\mathbb{R}^3}^{}\frac{\delta ^3(\mathbf{r})}{\left \| \mathbf{r-r'} \right \|}d^3\mathbf{r}-\int_{0}^{2\pi }d\phi \int_{0}^{\pi }\sin\theta d\theta \int_{0}^{\infty }\frac{\lambda ^2e^{-\lambda r'}}{r'\left \| \mathbf{r-r'} \right \|}r'^2dr' \right ]$$
$$\int_{\mathbb{R}^3}^{}\frac{\delta ^3(\mathbf{r})}{\left \| \mathbf{r-r'} \right \|}d^3\mathbf{r}=\frac{1}{\left \| \mathbf{r} \right \|}=\frac{1}{r}$$
In order to tackle the second integral I tried to use the law of cosines and stipulate that ##\mathbf{r}## is directed along the z-axis. I included the pictures of my work.
$$\int_{0}^{2\pi }d\phi \int_{0}^{\pi }\sin\theta d\theta \int_{0}^{\infty }\frac{\lambda ^2e^{-\lambda r'}}{r'\left \| \mathbf{r-r'} \right \|}r'^2dr'\rightarrow \int_{0}^{2\pi }d\phi\int_{0}^{\infty }\int_{0}^{\pi }\frac{\lambda ^2r'e^{-\lambda r'}}{(r^2+r'^2-2rr'\cos\theta )^{1/2}}\sin\theta d\theta dr'$$
Given that ##\int_{0}^{2\pi }d\phi =2\pi ##
The problem was evaluation of the integral
$$\int_{0}^{\infty }\int_{0}^{\pi }\frac{\lambda ^2r'e^{-\lambda r'}}{(r^2+r'^2-2rr'\cos\theta )^{1/2}}\sin\theta d\theta dr'$$$$\int_{0}^{\infty }\int_{0}^{\pi }\frac{\lambda ^2r'e^{-\lambda r'}}{(r^2+r'^2-2rr'\cos\theta )^{1/2}}\sin\theta d\theta dr'=\int_{0}^{\infty }\lambda ^2r'e^{-\lambda r}\left [ \int_{0}^{\pi }\frac{\sin\theta }{(r^2+r'^2-2rr'\cos\theta )^{1/2}}d\theta \right ]dr'$$
Using the substitution ##u=r^2+r'^2-2rr'\cos\theta##, I was able to integrate with respect to ##\theta## and arrive at
$$\int_{0}^{\infty }\lambda ^2r'e^{-\lambda r'}\times \frac{1}{rr'}\left \{ \sqrt{r'^2+r^2+2rr'}-\sqrt{r'^2+r^2-2rr'} \right \}dr'$$
$$\frac{\lambda ^2}{r}\left [ \int_{0}^{\infty }e^{-\lambda r'}\sqrt{r'^2+r^2+2rr'}dr'-\int_{0}^{\infty }e^{-\lambda r'}\sqrt{r'^2+r^2-2rr'}dr' \right ]$$
Note that I know for sure that the desired potential is
$$V(\mathbf{r})=A\frac{e^{-\lambda r}}{r}$$
However, the last integrals I arrived at, as far as I know, don't yield the answer I'm looking for. The exponential factor ends up canceling out entirely.
$$Q=\int_{\mathbb{R}^3}^{}\rho \, d^3\mathbf{r}$$
I found that the total charge of the configuration was zero.
Part (b) is where the difficulties arise for me. I applied
$$V(\mathbf{r})=\frac{1}{4\pi \epsilon _0}\int_{\Gamma }^{}\frac{\rho (\mathbf{r'})}{\left \| \mathbf{r-r'} \right \|}d^3\mathbf{r'}$$
to the given charge distribution. Here's my work.
$$V(\mathbf{r})=\frac{1}{4\pi \epsilon_0}\epsilon _0A\left [ 4\pi \int_{\mathbb{R}^3}^{}\frac{\delta ^3(\mathbf{r})}{\left \| \mathbf{r-r'} \right \|}d^3\mathbf{r}-\int_{0}^{2\pi }d\phi \int_{0}^{\pi }\sin\theta d\theta \int_{0}^{\infty }\frac{\lambda ^2e^{-\lambda r'}}{r'\left \| \mathbf{r-r'} \right \|}r'^2dr' \right ]$$
$$\int_{\mathbb{R}^3}^{}\frac{\delta ^3(\mathbf{r})}{\left \| \mathbf{r-r'} \right \|}d^3\mathbf{r}=\frac{1}{\left \| \mathbf{r} \right \|}=\frac{1}{r}$$
In order to tackle the second integral I tried to use the law of cosines and stipulate that ##\mathbf{r}## is directed along the z-axis. I included the pictures of my work.
$$\int_{0}^{2\pi }d\phi \int_{0}^{\pi }\sin\theta d\theta \int_{0}^{\infty }\frac{\lambda ^2e^{-\lambda r'}}{r'\left \| \mathbf{r-r'} \right \|}r'^2dr'\rightarrow \int_{0}^{2\pi }d\phi\int_{0}^{\infty }\int_{0}^{\pi }\frac{\lambda ^2r'e^{-\lambda r'}}{(r^2+r'^2-2rr'\cos\theta )^{1/2}}\sin\theta d\theta dr'$$
Given that ##\int_{0}^{2\pi }d\phi =2\pi ##
The problem was evaluation of the integral
$$\int_{0}^{\infty }\int_{0}^{\pi }\frac{\lambda ^2r'e^{-\lambda r'}}{(r^2+r'^2-2rr'\cos\theta )^{1/2}}\sin\theta d\theta dr'$$$$\int_{0}^{\infty }\int_{0}^{\pi }\frac{\lambda ^2r'e^{-\lambda r'}}{(r^2+r'^2-2rr'\cos\theta )^{1/2}}\sin\theta d\theta dr'=\int_{0}^{\infty }\lambda ^2r'e^{-\lambda r}\left [ \int_{0}^{\pi }\frac{\sin\theta }{(r^2+r'^2-2rr'\cos\theta )^{1/2}}d\theta \right ]dr'$$
Using the substitution ##u=r^2+r'^2-2rr'\cos\theta##, I was able to integrate with respect to ##\theta## and arrive at
$$\int_{0}^{\infty }\lambda ^2r'e^{-\lambda r'}\times \frac{1}{rr'}\left \{ \sqrt{r'^2+r^2+2rr'}-\sqrt{r'^2+r^2-2rr'} \right \}dr'$$
$$\frac{\lambda ^2}{r}\left [ \int_{0}^{\infty }e^{-\lambda r'}\sqrt{r'^2+r^2+2rr'}dr'-\int_{0}^{\infty }e^{-\lambda r'}\sqrt{r'^2+r^2-2rr'}dr' \right ]$$
Note that I know for sure that the desired potential is
$$V(\mathbf{r})=A\frac{e^{-\lambda r}}{r}$$
However, the last integrals I arrived at, as far as I know, don't yield the answer I'm looking for. The exponential factor ends up canceling out entirely.
