A parallel-plate capacitor is constructed from circular discs, each with a radius of 5.0 cm . Determine the capacitance if the separation between the plates is 1.0 mm .

Correct answer: D. 6.95 × 10^ -11 F Given: Radius of circular disc, r = 5.0 cm = 5 × 10^ -2 m Separation between plates, d = 1.0 mm = 10^ -3 m Permittivity of free space, _0 = 8.85 × 10^ -12 F/m Area of the plate, A = π r^2 = π × (5 × 10^ -2 )^2 = 25π × 10^ -4 m^2 Capacitance of a parallel-plate capacitor is given by: C = ( _0 A)/(d) Substituting the values: C = 8.85 × 10^ -12 × 25π × 10^ -4 10^ -3 C = 8.85 × 25 × 3.14 × 10^ -13 C = 694.7 × 10^ -13 F C 6.95 × 10^ -11 F

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Capacitors

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Capacitors

A parallel-plate capacitor is constructed from circular discs, each...

A parallel-plate capacitor is constructed from circular discs, each with a radius of

5.0 \text{ cm}

. Determine the capacitance if the separation between the plates is

1.0 \text{ mm}

Options

A $2.78 \times 10^{-10} \text{ F}$
B $6.95 \times 10^{-10} \text{ F}$
C $1.39 \times 10^{-10} \text{ F}$
D $6.95 \times 10^{-11} \text{ F}$

Correct answer

D. $6.95 \times 10^{-11} \text{ F}$

Step-by-step solution

Given:
Radius of circular disc,

r = 5.0 \text{ cm} = 5 \times 10^{-2} \text{ m}

Separation between plates,

d = 1.0 \text{ mm} = 10^{-3} \text{ m}

Permittivity of free space,

\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}

Area of the plate,

A = \pi r^2 = \pi \times (5 \times 10^{-2})^2 = 25\pi \times 10^{-4} \text{ m}^2

Capacitance of a parallel-plate capacitor is given by:

C = \dfrac{\epsilon_0 A}{d}

Substituting the values:

C = \dfrac{8.85 \times 10^{-12} \times 25\pi \times 10^{-4}}{10^{-3}}

C = 8.85 \times 25 \times 3.14 \times 10^{-13}

C = 694.7 \times 10^{-13} \text{ F}

C \approx 6.95 \times 10^{-11} \text{ F}

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