Show that the capacitance of the cylindrical capacitor shown in Fig. 23.3a approaches the capacitance of a parallel-plate capacitor if the separation d between the two cylinders is very small.

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Accepted Answer

When d = b − a is very small, then d/a must also be very small. If we use the approximation ln (1 + x) ≈ x for x << 1, in the natural logarithm of the denominator of Eq. 23.7, we find that:

[latex]{{{C={\frac{\ell}{2k\ln{(b/a)}}}={{{2\pi}}}\epsilon_{\mathrm{0}}\frac{\ell}{\ln{(b/a)}}}}}[/latex] (Cylindrical capacitor) (23.7)

[latex]\ln\left({\frac{b}{a}}\right)=\ln\left({\frac{a+d}{a}}\right)=\ln\left(1+{\frac{d}{a}}\right)\approx{\frac{d}{a}}\;\;\;\mathrm{(When}\;d/a\ll1)[/latex]

Then, using the surface area of the inner cylinder A = 2πa[latex]\ell[/latex], we find that Eq. 23.7 approaches Eq. 23.5 as follows:

[latex]C={\frac{\epsilon_{o}A}{d}}[/latex] (Parallel-plate capacitor) (23.5)

[latex]C=2\pi\epsilon_{o}{\frac{\ell}{\ln\left(b/a\right)}}\approx2\pi\epsilon_{o}{\frac{\ell}{d/a}}=\epsilon_{o}{\frac{2\pi a\ell}{d}}={\frac{\epsilon_{o}A}{d}}[/latex]

Question 23.2: Show that the capacitance of the cylindrical capacitor shown......

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