You are asked to construct a capacitor having a capacitance near1nFand a breakdown potential in excess of 10000 V . You think of using the sides of a tall Pyrex drinking glass as a dielectric, lining the inside and outside curved surfaces with aluminum foil to act as the plates. The glass is 15 cm tall with an inner radius of 3.6 cmand an outer radius of3.8 cm(a) What is the capacitance? (b) What is the breakdown potential of this capacitor?

a) Near capacitance, C = 1nF

b) Dielectric constant of Pyrex, k = 4.7

c) Inner radius of the glass, a = 3.6 cm

d) Outer radius of the glass, b = cm

e) Length of the glass, L = 15 cm

f) Present excess voltage that is breakdown voltage,$V_B={10}^{4}V$

g) Dielectric strength of Pyrex, $E=14\times {10}^{6}V/m$

The formula for the capacitance of a cylindrical capacitor can be used to find the required capacitance with the given geometry. Now, for the breakdown voltage, the dielectric strength of the material is taken. This determines the breakdown potential within the inner and outer radius of the cylinder.

Formulae:

The capacitance formula of a cylindrical surface with dielectric filled,

$C=k.\frac{2\mathrm{\pi }{\stackrel{,}{\mathrm{o}}}_0\mathrm{L}}{\ln \left({\displaystyle \frac{b}{a}}\right)}$ …(i)

The electric field strength relation to the voltage, E = V/d …(ii)

Where,

${\epsilon }_0=8.85\times {10}^{-12}\frac{C^2}{N.m^2}$ is permittivity of free space

Using the given data in equation (i), we can get the capacitance value between the plates with Pyrex field as follows:

$$\begin{gathered} C_1=\frac{4.7\times 2\mathrm{\pi }\times \left(8.85\times {10}^{-12}{\displaystyle \frac{{\mathrm{C}}^2}{\mathrm{N},{\mathrm{m}}^2}}\right)\times \left(15\times {10}^{-2}\mathrm{m}\right)}{\ln \left({\displaystyle \frac{3.8cm}{3.6cm}}\right)} \\ =0.73nF \end{gathered}$$

Hence, the value of the capacitance is $0.73nF$.

For Pyrex glass dielectric strength is$14\times {10}^{6}V/m$.

Breakdown potential of capacitor is given using equation (ii) as follows:

$$\begin{gathered} V_b=14\times {10}^{6}V/m.\left(3.8-3.6\right)\times {10}^{-2}m \\ =28kV \end{gathered}$$

Hence, the value of the voltage is 28 kV .