Question

A parallel plate capacitor is made of two rectangular plates of area 1.0 m^2 and the
separation between plates is 2.0 mm. A voltage of 100 V is applied across the
plates. Calculate the capacitance of the capacitor if a dielectric material of dielectric
constant 3.0 is introduced between the plates. What is the charge stored on each
plate of the capacitor?

Answer 1

The expression for the capacitance of a plane capator can be written as:

$$C = \frac{\varepsilon \varepsilon_0 S}{d},$$

where S is the area of the plates, d is the separation between the plates, epsilon is the dielectric material constant, epsilon₀ is the electric constant.

Substituting the numerical values, we obtain:

$$C=\frac{8.85 \cdot 10^{-12} \cdot 3 \cdot 1}{2 \cdot 10^{-3}} \approx 13.3 \cdot 10^{-9} \, F = 13.3 \, nF$$

The charge stored on each plate can be calculated as:

$$q = CV$$

Substituting the numerical values, we obtain:

$$q = 13.3 \cdot 10^{-9} \cdot 10^{2} = 13.3 \cdot 10^{-7} \, C \approx 1.3 \, \mu C$$

Answer: 13.3 nF, 1.3 muC.