Chapter 6, Problem 6.1P

To determine

The ratio of outer radius $b$ to inner radius $a$ in terms of the ratio of plate separation $d$ to plate width $w.$

Answer to Problem 6.1P

The ratio of outer radius $b$ to inner radius $a$ in terms of the ratio of plate separation $d$ to plate width $w$ is $\frac{b}{a}=\exp \left(2\pi \frac{d}{w}\right)$.

Explanation of Solution

Calculation:

The ratio of outer radius $b$ to inner radius $a$ in terms of the ratio of plate separation $d$ to plate width $w$ is expressed in such a way that there are two structures, and they store the same energy at a given applied voltage.

If they store the same energy at a given applied voltage, then the capacitances of both structures are the same. So, the length of both structure are same and they contain the same dielectric, it means the relative permittivity of both structure are the same.

The capacitance of first structure is written as,

$C=\frac{2\pi \epsilon }{\ln \left(\frac{b}{a}\right)}$    ...... (1)

Here,

$\epsilon$ is the permittivity.

The capacitance of second structure is written as,

$C=\frac{\epsilon w}{d}$    ...... (2)

The capacitances of both structures are same, so equation (1) is equal to equation (2), therefore, expression is written as,

$\frac{2\pi \epsilon }{\ln \left(\frac{b}{a}\right)}=\frac{\epsilon w}{d}$    ...... (3)

The simplified form of equation (3) is written as,

$\frac{b}{a}=\exp \left(2\pi \frac{d}{w}\right)$

Conclusion:

Therefore, the ratio of outer radius $b$ to inner radius $a$ in terms of the ratio of plate separation $d$ to plate width $w$ is $\frac{b}{a}=\exp \left(2\pi \frac{d}{w}\right)$.