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.


1 Answer 1


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.


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