- #1
Marcus Nielsen
- 4
- 0
Member advised to use the homework template for posts in the homework sections of PF.
Hello Guys! This is my first post so bear with me. I am currently studying the basics of electrostatics using the textbook "Introduction to electrodynamics 3 edt. - David J. Griffiths". My problem comes when i try to solve problem 2.21.
Find the potential V inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infinity as your reference point. Compute the gradient of V in each region, and check that it yields the correct field. Sketch V(r).
Using the equation V(r) = \int\limits_O^r \boldsymbol{E} \cdot \mathrm{d} \boldsymbol{l} and Gauss law \oint \boldsymbol{E} \cdot \mathrm{d} \boldsymbol{a} = \frac{Q_{enc}}{\epsilon_0}, I can solve the problem and get the same answer as in this guide http://www.physicspages.com/2011/10/08/electric-potential-examples/ first example. But my problem appears when I want to use the formula V(\boldsymbol{r})= \frac{1}{4 \pi \epsilon_0} \int \frac{\rho(\boldsymbol{r'})}{\eta} \mathrm{d}\tau' where \boldsymbol{\eta} = \boldsymbol{r}-\boldsymbol{r}' as defined in Griffiths textbook.
This is my calculation using the last formula.
We got a solid sphere with radius R and total charge q, therefore \rho = \frac{q}{\frac{4}{3} \pi R^3}
<br /> \begin{align*}<br /> V &= \frac{1}{4 \pi \epsilon_0} \int \frac{\rho(\boldsymbol{r'})}{\eta} \mathrm{d}\tau' \\<br /> &= \frac{1}{4 \pi \epsilon_0} \int\limits^{2 \pi}_0 \int\limits^{\pi}_{0} \int\limits^{R}_0 \frac{\rho}{\sqrt{r^2+z^2-2rz \cos(\theta)}} \sin(\theta) \ \mathrm{d}r \ \mathrm{d}\theta \ \mathrm{d}\phi<br /> \end{align*}<br />
Using substitution <br /> g =r^2+z^2-2rz \cos(\theta) \longrightarrow \mathrm{d}\theta = \frac{1}{2rz \sin(\theta)} \mathrm{d}g<br />
<br /> \begin{align*}<br /> V &= \frac{1}{4 \pi \epsilon_0} \int\limits^{2 \pi}_0 \int\limits^{g(\pi)}_{g(0)} \int\limits^{R}_0 \frac{\rho}{2rz\sqrt{g}} \ \mathrm{d}r \ \mathrm{d}g \ \mathrm{d}\phi\\<br /> &= \frac{1}{4 \pi \epsilon_0} 2 \pi \int\limits^R_0 \int\limits^{g(\pi)}_{g(0)} \frac{\rho}{2rz \sqrt{g}} \ \mathrm{d}g \ \mathrm{d}r\\<br /> &= \frac{1}{4 \epsilon_0} \int\limits^R_0 \int\limits^{g(\pi)}_{g(0)} \frac{\rho}{rz \sqrt{g}} \ \mathrm{d}g \ \mathrm{d}r\\<br /> &= \frac{1}{4 \epsilon_0} \int\limits^R_0 \frac{2\rho}{rz} \Big[ \sqrt{g} \Big]^{g(\pi)}_{g(0)} \ \mathrm{d}r\\<br /> &= \frac{1}{4 \epsilon_0} \int\limits^R_0 \frac{2\rho}{rz} \Big( \sqrt{r^2+z^2+2rz} - \sqrt{r^2+z^2-2rz} \Big) \ \mathrm{d}r\\<br /> &= \frac{1}{4 \epsilon_0} \int\limits^R_0 \frac{2\rho}{rz} \Big( \sqrt{(r+z)^2} - \sqrt{(r-z)^2} \Big) \ \mathrm{d}r\\<br /> \end{align*}<br />
At this stage I am a bit confused, the r will cancel out and i will be left with 1/r which will turn out to ln(r) after integration, does anyone know what am I doing wrong?
Find the potential V inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infinity as your reference point. Compute the gradient of V in each region, and check that it yields the correct field. Sketch V(r).
Using the equation V(r) = \int\limits_O^r \boldsymbol{E} \cdot \mathrm{d} \boldsymbol{l} and Gauss law \oint \boldsymbol{E} \cdot \mathrm{d} \boldsymbol{a} = \frac{Q_{enc}}{\epsilon_0}, I can solve the problem and get the same answer as in this guide http://www.physicspages.com/2011/10/08/electric-potential-examples/ first example. But my problem appears when I want to use the formula V(\boldsymbol{r})= \frac{1}{4 \pi \epsilon_0} \int \frac{\rho(\boldsymbol{r'})}{\eta} \mathrm{d}\tau' where \boldsymbol{\eta} = \boldsymbol{r}-\boldsymbol{r}' as defined in Griffiths textbook.
This is my calculation using the last formula.
We got a solid sphere with radius R and total charge q, therefore \rho = \frac{q}{\frac{4}{3} \pi R^3}
<br /> \begin{align*}<br /> V &= \frac{1}{4 \pi \epsilon_0} \int \frac{\rho(\boldsymbol{r'})}{\eta} \mathrm{d}\tau' \\<br /> &= \frac{1}{4 \pi \epsilon_0} \int\limits^{2 \pi}_0 \int\limits^{\pi}_{0} \int\limits^{R}_0 \frac{\rho}{\sqrt{r^2+z^2-2rz \cos(\theta)}} \sin(\theta) \ \mathrm{d}r \ \mathrm{d}\theta \ \mathrm{d}\phi<br /> \end{align*}<br />
Using substitution <br /> g =r^2+z^2-2rz \cos(\theta) \longrightarrow \mathrm{d}\theta = \frac{1}{2rz \sin(\theta)} \mathrm{d}g<br />
<br /> \begin{align*}<br /> V &= \frac{1}{4 \pi \epsilon_0} \int\limits^{2 \pi}_0 \int\limits^{g(\pi)}_{g(0)} \int\limits^{R}_0 \frac{\rho}{2rz\sqrt{g}} \ \mathrm{d}r \ \mathrm{d}g \ \mathrm{d}\phi\\<br /> &= \frac{1}{4 \pi \epsilon_0} 2 \pi \int\limits^R_0 \int\limits^{g(\pi)}_{g(0)} \frac{\rho}{2rz \sqrt{g}} \ \mathrm{d}g \ \mathrm{d}r\\<br /> &= \frac{1}{4 \epsilon_0} \int\limits^R_0 \int\limits^{g(\pi)}_{g(0)} \frac{\rho}{rz \sqrt{g}} \ \mathrm{d}g \ \mathrm{d}r\\<br /> &= \frac{1}{4 \epsilon_0} \int\limits^R_0 \frac{2\rho}{rz} \Big[ \sqrt{g} \Big]^{g(\pi)}_{g(0)} \ \mathrm{d}r\\<br /> &= \frac{1}{4 \epsilon_0} \int\limits^R_0 \frac{2\rho}{rz} \Big( \sqrt{r^2+z^2+2rz} - \sqrt{r^2+z^2-2rz} \Big) \ \mathrm{d}r\\<br /> &= \frac{1}{4 \epsilon_0} \int\limits^R_0 \frac{2\rho}{rz} \Big( \sqrt{(r+z)^2} - \sqrt{(r-z)^2} \Big) \ \mathrm{d}r\\<br /> \end{align*}<br />
At this stage I am a bit confused, the r will cancel out and i will be left with 1/r which will turn out to ln(r) after integration, does anyone know what am I doing wrong?