- #1
Tony Hau
- 101
- 30
Summary:: So this is a question from Griffiths' book on electrodynamics. The topic is on electircal potential. I have come up with a solution to a problem; the solution is wrong but I cannot spot the mistake.
So here is the question:
My answer to this question is: $$\sigma( θ ) = \frac {\epsilon_0} {2R} \sum_{l=0}^\infty (2l+1)(-l-1)C_1P_l (cos \theta) \text { whereas the solution in the textbook is } \frac {\epsilon_0} {2R} \sum_{l=0}^\infty (2l+1)^{2}C_1P_l (cos \theta)$$
The step is as follows:
Potential outside the sphere is $$V_{out} =\sum_{l=0}^\infty \frac {B_l} {r^{l+1}} \ P_l (cos \theta) $$
Potential inside the sphere is equal to ##V_o(\theta)##, on the grounds that a metal conductor has a constant potential inside it.
At ##r=R##, $$V_{out} =\sum_{l=0}^\infty \frac {B_l} {R^{l+1}} \ P_l (cos \theta) = V_o(\theta) $$
Multiplying both sides by ##P_{l^{'}}cos(\theta)sin(\theta)d\theta## and integrate from ##0## to ##\pi##, we get $$\sum_{l=0}^\infty \frac {B_l} {R^{l+1}} \int_0^\pi P_{l^{'}}cos(\theta)P_{l}cos(\theta)sin(\theta)d\theta = \int_0^\pi V_o(\theta)P_{l^{'}}cos(\theta)sin(\theta)d\theta$$.
Because $$\int_0^\pi P_{l^{'}}cos(\theta)P_{l}cos(\theta)sin(\theta)d\theta =
\begin{cases}
0 & \text{if } l \neq l^{'} \\
\frac{2} {2l+1}\ & \text{if } l = l^{'}
\end{cases}
$$
$$B_{l} = \frac {(2l+1)R^{l+1}}{2}\ \int_0^\pi V_o(\theta)P_{l^{'}}cos(\theta)sin(\theta)d\theta$$.
Therefore, $$V_{out} =\sum_{l=0}^\infty \frac {(2l+1)R^{l+1}}{2r^{l+1}}\ \int_0^\pi V_o(\theta)P_{l^{'}}cos(\theta)sin(\theta)d\theta P_{l}cos(\theta)$$
$$ =\sum_{l=0}^\infty \frac {(2l+1)R^{l+1}}{2r^{l+1}}\ C_{1} P_{l}cos(\theta)$$
The charge density at the surface is given by: $$\sigma(\theta)\ = (\frac {\partial V_{out}} {\partial r} - \frac {\partial V_{in}} {\partial r})\epsilon_o \text{ }at \text{ } r=R$$
The derivative of ##V_{in}\text{ } is \text{ }0##. Therefore, $$\sigma(\theta)\ = \epsilon_o\sum_{l=0}^\infty \frac{\partial}{\partial r}\frac {(2l+1)R^{l+1}}{2r^{l+1}}\ C_{1} P_{l}cos(\theta)$$
$$= \epsilon_o\sum_{l=0}^\infty (-l-1)\frac {(2l+1)R^{l+1}}{2r^{l+2}}\ C_{1} P_{l}cos(\theta)$$
Therefore, at ##r=R##, $$\sigma(\theta)\ = \epsilon_o\sum_{l=0}^\infty (-l-1)\frac {(2l+1)R^{l+1}}{2R^{l+2}}\ C_{1} P_{l}cos(\theta)$$
$$ = \frac {\epsilon_0} {2R} \sum_{l=0}^\infty (2l+1)(-l-1)C_1P_l (cos \theta) $$
What is wrong with the steps?By the way, this is my first time using LaText. The process is time consuming and I do not understanding why simply uploading a picture of my steps is prohibited; I have to write in Latex, which has costed me more than an hour.[Moderator's note: Moved from a technical forum and thus no template.]
So here is the question:
My answer to this question is: $$\sigma( θ ) = \frac {\epsilon_0} {2R} \sum_{l=0}^\infty (2l+1)(-l-1)C_1P_l (cos \theta) \text { whereas the solution in the textbook is } \frac {\epsilon_0} {2R} \sum_{l=0}^\infty (2l+1)^{2}C_1P_l (cos \theta)$$
The step is as follows:
Potential outside the sphere is $$V_{out} =\sum_{l=0}^\infty \frac {B_l} {r^{l+1}} \ P_l (cos \theta) $$
Potential inside the sphere is equal to ##V_o(\theta)##, on the grounds that a metal conductor has a constant potential inside it.
At ##r=R##, $$V_{out} =\sum_{l=0}^\infty \frac {B_l} {R^{l+1}} \ P_l (cos \theta) = V_o(\theta) $$
Multiplying both sides by ##P_{l^{'}}cos(\theta)sin(\theta)d\theta## and integrate from ##0## to ##\pi##, we get $$\sum_{l=0}^\infty \frac {B_l} {R^{l+1}} \int_0^\pi P_{l^{'}}cos(\theta)P_{l}cos(\theta)sin(\theta)d\theta = \int_0^\pi V_o(\theta)P_{l^{'}}cos(\theta)sin(\theta)d\theta$$.
Because $$\int_0^\pi P_{l^{'}}cos(\theta)P_{l}cos(\theta)sin(\theta)d\theta =
\begin{cases}
0 & \text{if } l \neq l^{'} \\
\frac{2} {2l+1}\ & \text{if } l = l^{'}
\end{cases}
$$
$$B_{l} = \frac {(2l+1)R^{l+1}}{2}\ \int_0^\pi V_o(\theta)P_{l^{'}}cos(\theta)sin(\theta)d\theta$$.
Therefore, $$V_{out} =\sum_{l=0}^\infty \frac {(2l+1)R^{l+1}}{2r^{l+1}}\ \int_0^\pi V_o(\theta)P_{l^{'}}cos(\theta)sin(\theta)d\theta P_{l}cos(\theta)$$
$$ =\sum_{l=0}^\infty \frac {(2l+1)R^{l+1}}{2r^{l+1}}\ C_{1} P_{l}cos(\theta)$$
The charge density at the surface is given by: $$\sigma(\theta)\ = (\frac {\partial V_{out}} {\partial r} - \frac {\partial V_{in}} {\partial r})\epsilon_o \text{ }at \text{ } r=R$$
The derivative of ##V_{in}\text{ } is \text{ }0##. Therefore, $$\sigma(\theta)\ = \epsilon_o\sum_{l=0}^\infty \frac{\partial}{\partial r}\frac {(2l+1)R^{l+1}}{2r^{l+1}}\ C_{1} P_{l}cos(\theta)$$
$$= \epsilon_o\sum_{l=0}^\infty (-l-1)\frac {(2l+1)R^{l+1}}{2r^{l+2}}\ C_{1} P_{l}cos(\theta)$$
Therefore, at ##r=R##, $$\sigma(\theta)\ = \epsilon_o\sum_{l=0}^\infty (-l-1)\frac {(2l+1)R^{l+1}}{2R^{l+2}}\ C_{1} P_{l}cos(\theta)$$
$$ = \frac {\epsilon_0} {2R} \sum_{l=0}^\infty (2l+1)(-l-1)C_1P_l (cos \theta) $$
What is wrong with the steps?By the way, this is my first time using LaText. The process is time consuming and I do not understanding why simply uploading a picture of my steps is prohibited; I have to write in Latex, which has costed me more than an hour.[Moderator's note: Moved from a technical forum and thus no template.]
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