Answer:
[tex]C=1.12\times10^{-10}F[/tex]
Explanation:
The capacitance of a parallel-plate capacitor is given by the formula:
[tex]C=\frac{k \epsilon_0 A}{d}[/tex]
where k is the dielectric constant of the material between them, A the area of the plates, d the distance between them and [tex]\epsilon_0=8.85\times10^{-12}F/m[/tex] the permittivity, an universal constant.
Since the area of the plates is the product of their sides a=4.3cm=0.043m and b=4.2cm=0.042m, in S.I. we have:
[tex]C=\frac{k \epsilon_0 ab}{d}=\frac{(2.1)(8.85\times10^{-12}F/m)(0.043m)(0.042m)}{(0.0003m)}=1.12\times10^{-10}F[/tex]
Capacitance is the effect of a capacitor. The capacitance of the parallel plate capacitor is 1.1193 x 10⁻¹⁰ F.
The capacitance is the effect of a capacitor, while a capacitor is a device that stores the electrical energy into it. The energy stored in a capacitor can be calculated by the formula,
[tex]U = \dfrac{1 \times Q^2}{2 \times C} = \dfrac{1 \times Q}{2 \times V} = \dfrac{1}{2}CV^2[/tex]
Given to us
Plate dimensions = 4.30cm by 4.20 cm
Area of the plate, A = 0.001806 m²
Distance between the two plates, d = 0.3 mm = 0.0003 m
Dielectric constant, k = 2.1
The permittivity of free space, ε₀ = 8.85 x 10-12 farad per meter (F/m)
We know that capacitance of a parallel plate capacitor is given as,
[tex]C = \dfrac{k\epsilon_0A}{d}[/tex]
Substitute the value,
[tex]C = \dfrac{2.1 \times 8.854 \times 10^{-12} \times 0.001806 }{0.0003}\\\\C = 1.1193 \times 10^{-10}\ F[/tex]
Hence, the capacitance of the parallel plate capacitor is 1.1193 x 10⁻¹⁰ F.
Learn more about Capacitance:
https://brainly.com/question/17773926
