What is the limit I have to take so that the capacitance formula for concentric cylinders shows the capacitance of infinite plates per surface area?
From Wikipeida,
$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots $$
where \$|x| &lt; 1\$.
For very large values of the cylinder diameter and small gap (\$d = b-a\$) between them,
$$\ln(\frac{b}{a}) = \ln(\frac{a+d}{a})=\ln(1+\frac{d}{a}) = \frac{d}{a} - \dots$$
\$|d/a| &lt; 1\$ for large diameter cylinder with small gap between plates.
Neglecting the higher powers, the formula given in the question for cylinder becomes
$$C \approx 2\pi \varepsilon L / (d/a) = 2\pi \varepsilon L \cdot a/ d$$
Note that

\$d\$ appears in the denominator as in target formula of parallel plates.
\$\varepsilon\$ appears in the numerator as in target formula.
\$2 \pi a \approx 2\pi b\$ is circumference of the inner / outer pipe.
\$2 \pi a L\$ is an area similar to that in the target formula.


I couldn't figure out how to eliminate the 2piL part.

With the above shown approximation of \$\ln()\$, it is clear that you need to group \$2\pi\$ with \$a\$ rather than \$L\$.

I have probably not gotten the actual concept of the problem in the first place.

The &quot;vertical&quot; direction of parallel plates correspond to L for the cylinder. The &quot;horizontal&quot; direction of the parallel plates correspond to the circumferential direction of the cylinder. A way to visualise this is that you &quot;cut&quot; the cylinder(s) along the length and unroll it to get flat rectangular plate(s). This rectangular plate(s) then look similar to the rectangular plates in the parallel plate capacitor.