Question

A parallel plate capacitor consists of two identical rectangular metal
plates of sides 5.5 cm and 4 cm which are separated by 0.7 mm. Calculate the capacitance of the
capacitor when the space is filled with (a) air and (b) a medium of dielectric constant 4.

Answer 1

Explanation:

separation d=0.07mm

d=0.7×10^-3×10^2

d=0.07cm

rectangular metal plates of sides 5.5 and 4

area of rectangle=length×breadth

area A=5.5×4

A =22 sq cm

capacitance C=€A/d

C=(8.84×10^-12)×22/0.07

C=27.78×10^-10 Fcm

Answer 2

The capacitance of the capacitor when the space is filled with

(a) air is 2.7826 × 10^(-11), and

(b) the medium of dielectric constant 4 is 11.1304 × 10^(-11)

Given,

Length of capacitor plate, l = 5.5 cm = 0.055 m

Breadth of capacitor plate, b = 4 cm = 0.04 m

Distance between the capacitor plates. d = 0.7 mm = 0.0007 m

Dielectric constant of the medium, k = 4

To Find,

The capacitance of the capacitor when the space is filled with

(a) air, and

(b) another medium

Solution,

A capacitor's electrical property, or capacitance, measures how well it can hold an electrical charge on its two plates.

The mathematical formula for capacitance is:

$\implies C = \epsilon \hspace{0.1 cm} \frac{A}{d}$

We have a parallel plate capacitor of two identical rectangular metal plates.

The length and breadth of the capacitor plates are given.

Area of the capacitor plate, A = l × b = 0.055 m × 0.04 m = 0.0022 m²

Also, given the distance between the plates, d = 0.0007 m

We have to find the capacitance of the the capacitor in two scenarios.

(a) when the space between the plates is filled with air:

∈ = ∈0 = 8.854 × 10^(-12)

The capacitance in this case is given by:

$\implies C_a = \epsilon_0 \hspace{0.1 cm} \frac{A}{d}\\\\\implies C_a = 8.854 \times 10^{-12} \times \frac{0.0022}{0.0007}\\\\\implies C_a = 8.854 \times 10^{-12} \times 3.1428\\\\\implies C_a = 8.854 \times 10^{-12} \times 3.1428\\\\\implies C_a = 2.7826 \times 10^{-11}$

(b) when the space between the plates is filled with a medium of dielectric constant 4:

∈ = k × ∈0 = 4 × 8.854 × 10^(-12)

The capacitance in this case is given by:

$\implies C_m = \epsilon \hspace{0.1 cm} \frac{A}{d}\\\\\implies C_m = 4 \times \epsilon_0 \hspace{0.1 cm} \frac{A}{d}\\\\\implies C_m = 4 \times C_a\\\\\implies C_m = 4 \times 2.7826 \times 10^{-11}\\\\\implies C_m = 11.1304 \times 10^{-11}$

Therefore, the capacitance of the capacitor when the space is filled with air is 2.7826 × 10^(-11) and the medium of dielectric constant 4 is 11.1304 × 10^(-11)

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