Question

1. A Cylindrical capacitor having a length of 8 cm is made of two concentric rings with an inner radius as 3 cm and outer radius as 6 cm. Find the capacitance of the capacitor.

2. A parallel-plate capacitor is filled with polystyrene of 0.0002 m thickness. Find the plate area if the new capacitance after the insertion is 3.4 µF.

Answer 1

1. The capacitance of a cylindrical capacitor is given as follows:

$$C = \dfrac{2\pi \epsilon_0 L}{\ln(r_2/r_1)}$$

where $L=8cm = 0.08m$, $r_1 = 0.03m, r_2 = 0.06m$ and $\epsilon_0 \approx 8.85\times 10^{-12} F/m$ is vacuum permittivity. Thus, obtain:

$$C = \dfrac{2\pi \cdot 8.85\times 10^{-12}\cdot 0.08}{\ln(0.06/0.03)} \approx 6.4\times 10^{-12}F$$

2. The capacitance of a parallel plate capacitor is:

$$C=\dfrac{\epsilon_0 \epsilon A}{d}$$

where $A$ is the plate area, $\epsilon =2.55$ is the dielectric constant of polystyrene, and $d = 0.0002 m$ is the distance between plates. Expressing $A$, get:

$$A = \dfrac{dC}{\epsilon_0 \epsilon} =\dfrac{0.0002m\cdot 3.4\times10^{-6}F }{8.85\times 10^{-12}F/m\cdot 2.55} \approx 30m^2$$