5
$\begingroup$

The electric potential of a charge between two infinite conducting planes can be expressed as an infinite series (of image charges) [Kellogg1929] as

$$ V=\sum_{n=-\infty}^{\infty}\left( \frac{1}{\sqrt{(z-2na-z_c)^2+x^2}} - \frac{1}{\sqrt{(z-2na+z_c)^2+x^2}} \right) $$ (for $a > 0$, $0 < z < a$, $0 < z_c < a$, and $x > 0$)

No closed form for this series seems to exist. Yet, when inspecting sums up to $n=100,000,000$ I found the result to closely follow

$$ V = \frac{P}{a}\sin\left(\frac{z}{a}\pi\right)\sin\left(\frac{z_c}{a}\pi\right)e^{-\frac{Q}{a}x} $$ in the range where $x > a$. The parameters I found empirically are $P=2.5$ and $Q=3.25$. The closed form is not equal to the infinite series but it comes close enough (it deviates by 10%-20% in absolute value over 12 orders of magnitude ($a < x < 10a$) that to my intuition there must be a way to show that the infinite series has a predominantly exponential behavior. Is that possible and how could that be shown? I understand that exponential functions are infinite power series, but the first series is not a power series.

$\endgroup$
6
$\begingroup$

This approximate solution can be obtained by transforming the infinite sum to a contour integral and then making an asymptotic expansion. This is worked out in Application of Sommerfeld-Watson Transformation to an Electrostatics Problem (1969). The result is slightly different from your surmise, $$V\approx\sqrt{\frac{8}{xa}}\sin(\pi z/a)\sin(\pi z_c/a)e^{-\pi x/a}\;\;\text{if}\;\;x\gtrsim a,$$ so your factor $Q=3.25$ in the exponential decay has the exact value $Q=\pi$, but more significantly there is a pre-exponential decay $\propto 1/\sqrt{x}$.

An alternative way to arrive at this asymptotic expression is to rewrite the sum over image charges identically as $$V=\frac{4}{a}\sum_{n=1}^\infty \sin(n\pi z/a)\sin(n\pi z_c/a)K_0(n\pi x/a)$$ and then make an asymptotic expansion of the Bessel function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.