Two infinitely-long grounded metal plates, again at y = 0 and y=a, are connected at x=±b by metal strips maintained at a constant potential V0, as shown in Fig.3.20(a thin layer of insulation at each corner prevents them from shorting out).Find the potential inside the resulting rectangular pipe.

Question

Two infinitely-long grounded metal plates, again at y = 0 and y = a, are connected at x = ±b by metal strips maintained at a constant potential [latex]V_{0}[/latex], as shown in Fig. 3.20 (a thin layer of insulation at each corner prevents them from shorting out). Find the potential inside the resulting rectangular pipe.

Accepted Answer

Once again, the configuration is independent of z. Our problem is to solve Laplace’s equation

[latex]\frac{\delta ^{2}V}{\delta x^{2}}+\frac{\delta ^{2}V}{\delta y^{2}}=0 [/latex]

subject to the boundary conditions

[latex]\begin{cases} (i) V = 0\quad when\quad y = 0, \(ii) V = 0\quad when\quad y = a,\(iii) V = V_{0}\quad when\quad x = b,\(iv) V = V_0\quad when\quad x = −b.\end{cases} [/latex] (3.40)

The argument runs as before, up to Eq. 3.27:

[latex]V(x,y)=(Ae^{kx}+Be^{-kx}) (C\sin ky + D\cos ky),[/latex] (3.27)

This time, however, we cannot set A = 0; the region in question does not extend to [latex]x =∞, so e_{kx}[/latex] is perfectly acceptable. On the other hand, the situation is symmetric with respect to x, so V(−x, y) = V(x, y), and it follows that A = B. Using

[latex]e^{kx} + e^{−kx} = 2 \cosh kx,[/latex]

and absorbing 2A into C and D, we have

[latex]V(x, y) = \cosh kx (C \sin ky + D \cos ky).[/latex]

Boundary conditions (i) and (ii) require, as before, that D = 0 and k = nπ/a, so

[latex]V(x, y) = C \cosh (nπx/a) \sin (nπy/a).[/latex] (3.41)

Because V(x, y) is even in x, it will automatically meet condition (iv) if it fits (iii). It remains, therefore, to construct the general linear combination,

[latex]V(x, y) =\sum\limits_{n=1}^{\infty }{C_{n} \cosh(nπx/a) \sin(nπy/a),} [/latex]

and pick the coefficients [latex]C_{n}[/latex] in such a way as to satisfy condition (iii):

[latex]V(b, y) =\sum\limits_{n=1}^{\infty }{C_{n} \cosh(nπx/a) \sin(nπy/a)} =V_{0}.[/latex]

This is the same problem in Fourier analysis that we faced before; I quote the result from Eq. 3.35:

[latex]C_{n}=\frac{2V_{0}}{a} \int_{0}^{a}{} \sin (n \pi y/a)=\frac{2V_{0}}{n\pi }(1-\cos n\pi ) =\begin{cases} 0, if\quad n\quad is\quad even, \ \frac{a}{2}, if\quad n\quad is\quad odd\end{cases}. [/latex] (3.35)

[latex]C_{n}\cosh (n\pi b/a)=\begin{cases}0, if\quad n\quad is\quad even\\frac{4V_{0}}{n\pi}, if\quad n\quad is\quad odd\end{cases} [/latex]

Conclusion: The potential in this case is given by

[latex]V(x,y)=\frac{4V_{0}}{\pi}\sum\limits_{n=1.3.5....}^{}{\frac{1}{n}\frac{\cosh (n\pi x/a)}{\cosh (n\pi b/a)}\sin(n \pi y/a) } . [/latex] (3.42)

This function is shown in Fig. 3.21.

Question 3.4: Two infinitely-long grounded metal plates, again at y = 0 an...

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