Question

The parallel plates of capacitors are 5 cm x 10 cm and have a separation of 2cm. If the space between them the plates is filled with bakalite (k=4.7) what is the capacitance?

Answer 1

The capacitance of a parallel plate capacitor is given as follows:

$$C = \dfrac{\varepsilon_0\varepsilon A}{d}$$

where $\varepsilon_0 = 8.85\times 10^{-12}F/m$ is the vacuum permittivity, $\varepsilon = 4.7$ is the bakalite dielectric permittivity, $A = 5cm\times 10cm = 50cm^2 = 0.005m^2$ is the area of one plate, and $d = 2cm = 0.02m$ is the distance between the plates. Thus, obtain:

$$C = \dfrac{8.85\times 10^{-12}F/m\cdot 4.7 \cdot 0.005m^2}{0.02m} \approx 1.04\times 10^{-12}F$$

Answer. $1.04\times 10^{-12}F$.