Question 6.6: Find the potential distribution in Example 6.5 if Vo is not ...

(a) In Example 6.5, every step before eq. (6.5.19) remains the same; that is, the solution is of the form

V(x,y)=\sum\limits_{n=1}^{\infty}c_{n}\sin\frac{n\pi x}{b}\sinh\frac{n\pi y}{b}                                                                        (6.6.1)

in accordance with eq. (6.5.18).

V(x,y)=\sum\limits_{n=1}^{\infty}c_{n}\sin\frac{n\pi x}{b}\sinh\frac{n\pi y}{b}

But instead of eq. (6.5.19)

V(x,y=a)=V_{o}=\sum\limits_{n=1}^{\infty}c_{n}\sin\frac{n\pi x}{b}\sinh\frac{n\pi a}{b}

we now have

V(y=a)=V_{o}=10\sin\frac{3\pi x}{b}=\sum\limits_{n=1}^{\infty}c_{n}\sin\frac{n\pi x}{b}\sinh\frac{n\pi a}{b}

By equating the coefficients of the sine terms on both sides, we obtain

c_{n}=0,       n\neq 3

For n=3

10=c_{3}\sinh\frac{3\pi a}{b}

or

c_{3}=\frac{10}{\sinh\frac{3\pi a}{b}}

Thus the solution in eq. (6.6.1) becomes

V(x,y)=10\sin\frac{3\pi x}{b}\frac{\sinh\frac{3\pi y}{b}}{\sinh\frac{3\pi a}{b}}

(b) Similarly, instead of eq. (6.5.19), we have

V_{o}=V(y=a)

or

2\sin\frac{\pi x}{b}+\frac{1}{10}\sin\frac{5\pi x}{b}=\sum\limits_{n=1}^{\infty}c_{n}\sin\frac{n\pi x}{b}\sinh\frac{n\pi a}{b}

Equating the coefficient of the sine terms:

c_{n}=0,       n\neq 1,5

For n=1

2=c_{1}\sinh\frac{\pi a}{b}   or  c_{1}=\frac{2}{\sinh\frac{\pi a}{b}}

For n=5

\frac{1}{10}=c_{5}\sinh\frac{5\pi a}{b}   or   c_{5}=\frac{1}{10\sinh\frac{5\pi a}{b}}

Hence

V(x,y)=\frac{2\sin\frac{\pi x}{b}\sinh\frac{\pi y}{b}}{\sinh\frac{\pi a}{b}}+\frac{\sin\frac{5\pi x}{b}\sinh\frac{5\pi y}{b}}{10\sinh\frac{5\pi a}{b}}