Question

A parallel plate capacitor having a plate area of 0.70 m2 and a plate separation of 1.0 mm is connected to a source with a voltage of 50 V. Find the capacitance, the charge on the plates, and the energy of the capacitor b) when the capacitor contains polystyrene

Answer 1

1. The capacitance of the parallel plate capacitor is given as follows:

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

where $\varepsilon_0 = 8.85 \cdot 10^{−12}F/m$ is the electric constant, $\varepsilon = 2.5$ is the permittivity of polystyrene, $S = 0.7m^2$ is the area of the plates, and $d = 1mm = 10^{-3}m$ is the distance between the plates.

Thus, obtain:

$$C = \dfrac{8.85 \cdot 10^{−12}\cdot 2.5\cdot 0.7}{10^{-3}} \approx 15.5\times10^{-9}F$$

2. The charge on the plates (on one plate) is:

$$q = CV$$

where $V = 50V$ is the voltage of the source. Thus, obtain:

$$q = 15.5\times 10^{-9}F\cdot 50V \approx 7.8\times 10^{-7}C$$

3. The energy stored in the capacitors is:

$$W = \dfrac{CV^2}{2}\\ W = \dfrac{15.5\times10^{-9}\times 50^2}{2}\approx 1.9\times 10^{-5}J$$

Answer. a) $15.5\times10^{-9}F$, b) $7.8\times 10^{-7}C$, c) $1.9\times 10^{-5}J$.