Chapter 24 Problem 39

Problem

The capacitor, coordinate system, and the infinitesimal elements

• A parallel-plate capacitor is tilted so it makes a small angle $\theta$ with the other plate.
• capacitance C in terms of A, d, and $\theta$, where A is the area of each plate
• $\theta$ is small.
• Assume the plates are square.

Solution

The small angle approximation makes this problem much easier.

The distance between the plates is

$d+x=d+t\tan \theta \approx d+y\theta$

The plates are square, thus the length of the edge should be

$A=l^{2}\rightarrow l={\sqrt {A}}$

then the infinitesimal area that we will integrate is

$dA={\sqrt {A}}dy$

since $\cos \theta \approx 1$ in small angle approximation, these are parallel plate capacitors connected in parallel.

$dC=\epsilon _{0}{\frac {dA}{d+y\theta }}=\epsilon _{0}{\frac {{\sqrt {A}}dy}{d+y\theta }}$

$C=\int dC=\int _{0}^{\sqrt {A}}\epsilon _{0}{\frac {{\sqrt {A}}dy}{d+y\theta }}$

$u=d+y\theta \rightarrow du=\theta dy$

$C=\left.{\frac {\epsilon _{0}{\sqrt {A}}}{\theta }}\ln(d+y\theta )\right|_{0}^{\sqrt {A}}={\frac {\epsilon _{0}{\sqrt {A}}}{\theta }}\left[\ln(d+\theta {\sqrt {A}})-\ln(d)\right]$ $={\frac {\epsilon _{0}{\sqrt {A}}}{\theta }}\ln \left({\frac {d+\theta {\sqrt {A}}}{d}}\right)={\frac {\epsilon _{0}{\sqrt {A}}}{\theta }}\ln \left(1+{\frac {\theta {\sqrt {A}}}{d}}\right)$

$\ln(1+x)=f(a)+{\frac {1}{1+a}}{\frac {x-a}{1!}}-{\frac {1}{(1+a)^{2}}}{\frac {(x-a)^{2}}{2!}}+{\frac {2}{(1+a)^{3}}}{\frac {(x-a)^{3}}{3!}}-{\frac {6}{(1+a)^{4}}}{\frac {(x-a)^{4}}{4!}}+...$

Since $\theta$ is small, the Maclaurin expansion (a=0) reads $\ln(1+x)\approx x-{\frac {1}{2}}x^{2}$

then

$C={\frac {\epsilon _{0}{\sqrt {A}}}{\theta }}\left[{\frac {\theta {\sqrt {A}}}{d}}-{\frac {1}{2}}\left({\frac {\theta {\sqrt {A}}}{d}}\right)^{2}\right]={\frac {\epsilon _{0}A}{d}}\left(1-{\frac {\theta {\sqrt {A}}}{2d}}\right)$