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I'm currently delving into the fascinating topic of electrostatics, specifically the distribution of potential in configurations involving conducting plates and charged wires. My focus is on a setup consisting of two infinite conducting plates, one aligned along the x-axis and the other positioned at $y=a$, with an infinite conducting wire carrying a charge $q$ placed between them at $y=d$ (where $d<a$), extending along the $z$-axis. The charge density $\sigma$) on the wire is described by $\sigma = q\delta(y-d)$. This system is characterized by differing potentials on either side of the x-axis: $\phi_1$ for $x<0$ and $\phi_2$ for $x>0$, with the potential $\phi$ being zero at $y=0$ and $y=a$. An intriguing boundary condition links the charge density to the potential's spatial derivatives, as

$$\frac{\sigma}{\epsilon_0}=\frac{\partial\phi_1 }{\partial x}-\frac{\partial\phi_2 }{\partial x}.$$

In an attempt to better understand the underlying physics and mathematical descriptions of potential distributions in such scenarios, I've been exploring solutions to the Laplace equation for this configuration. Using the separation of variables, I arrived at a general solution form:

$$\phi(x,y)=(A\sin(ky)+B\cos(ky))(Ce^{-kx}+De^{kx})\text{where}\; k=\frac{n\pi}{a}.$$

Integrating further, especially when considering the discontinuity jump boundary condition and applying the orthogonality principle, I encountered a conceptual hurdle related to integrating functions involving $\sin\left(\frac{n\pi d}{a}\right)$, which appears as a constant in the context of the integral:

\begin{align}\int_{0}^{a} \frac{q}{\epsilon_0} \sin\left(\frac{m\pi y}{a}\right) dy =& \int_{0}^{a} \left[ \sum_{n} -\frac{n\pi}{a} A_n \sin\left(\frac{n\pi d}{a}\right) \sin\left(\frac{m\pi y}{a}\right)\right. \\& +\left. \sum_{n} \frac{n\pi}{a} B_n \sin\left(\frac{n\pi d}{a}\right) \sin\left(\frac{m\pi y}{a}\right) \right] dy\end{align}

This brings me to my conceptual query: How does the presence of $\sin\left(\frac{n\pi d}{a}\right)$, which is essentially a constant within the integral, affect the application of the orthogonality condition in this context? I'm interested in understanding the theoretical approach and the principles at play when dealing with such boundary conditions and their implications on potential distributions, rather than seeking a direct solution to this problem.

Any insights into the principles governing the application of orthogonality in the context of electrostatics, especially with such specific boundary conditions, would be greatly appreciated.

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  • $\begingroup$ I tried to redo the calculations but I cannot figure out how you got your last equation from the boundary condition. $\endgroup$
    – Christophe
    Mar 20 at 14:20
  • $\begingroup$ @Christophe I have omitted this part. We know that at the $(0,d)$ we have the boundary condition: $$\frac{\sigma}{\epsilon_0}=\frac{\partial\phi_1 }{\partial x}-\frac{\partial\phi_2 }{\partial x}.$$ In here the exponential terms vanish after getting the partial derivatives. Then I've applied the orthogonality condition. $\endgroup$
    – SaaN
    Mar 20 at 19:10

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Using the fact that the potential is symmetric under the exchange $x\rightarrow -x$, should vanish for $y=0$ and $y=a$ and cannot diverge in the limit $|x|\rightarrow +\infty$, the solution of Laplace equation is restricted to $$\varphi(x,y)=\sum_{n=1}^{+\infty} B_n\sin {n\pi y\over a}e^{-{n\pi |x|\over a}}$$ The plan $y=0$ carries a charge density $$\sigma(y,z)=\lambda\delta(y-d)$$ which leads to the condition $$-\left({\partial\varphi\over\partial x}\right)_{(0^+,y,z)} +\left({\partial\varphi\over\partial x}\right)_{(0^-,y,z)} ={\sigma\over\varepsilon_0}={\lambda\over\varepsilon_0}\delta(y-d)$$ Plugging the solution, we get $$\sum_{n=1}^{+\infty} 2B_n{n\pi\over a}\sin {n\pi y\over a} ={\lambda\over\varepsilon_0}\delta(y-d)$$ Mulitplying by $\sin {m\pi y/a}$ and then integrating over $[0;a]$ $$\sum_{n=1}^{+\infty} 2B_n{n\pi\over a} \int_0^a \sin {n\pi y\over a}\sin {m\pi y\over a}dy ={\lambda\over\varepsilon_0}\sin {m\pi d\over a}$$ The integral equals ${a\over 2}\delta_{n,m}$. Finally, the potential is $$\varphi(x,y)={\lambda\over\pi\varepsilon_0}\sum_{n=1}^{+\infty} {1\over n}\sin {n\pi d\over a}\sin {n\pi y\over a}e^{-{n\pi|x|\over a}}$$

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  • $\begingroup$ I got what you have said here. But what is $\lambda$? $\endgroup$
    – SaaN
    Mar 20 at 19:14
  • $\begingroup$ Sorry, I denoted $\lambda$ the linear charge density on the wire. I usually reserve $q$ for a point charge. $\endgroup$
    – Christophe
    Mar 20 at 19:52
  • $\begingroup$ In this situation, we assume that $q$ amount of charge is spread over the conducting wire. Also, the wire is in the $z$ direction and the problem is $2$ dimensional we can consider the linear charge density as a point charge in the $2D$ situation. That's why I put $q$ there. $\endgroup$
    – SaaN
    Mar 21 at 5:05

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