A parallel-plate capacitor having a plate area of 20 cm^2 and a plate separation of 1.0 mm is connected to a battery. The circuit has a resistance of 10 k . Determine the time constant of the circuit.

Correct answer: C. 0.18 The capacitance of a parallel-plate capacitor is given by C = ( _0 A)/(d) Substituting the given values: C = 8.85 × 10^ -12 × 20 × 10^ -4 1.0 × 10^ -3 C = 17.7 × 10^ -12 F The time constant of the RC circuit is given by = RC = 10 × 10^3 × 17.7 × 10^ -12 = 17.7 × 10^ -8 s = 0.177 0.18

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Electric Current in Conductors

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Current in Conductors

A parallel-plate capacitor having a plate area of 20 cm^2 and a...

A parallel-plate capacitor having a plate area of

20 \text{ cm}^2

and a plate separation of

1.0 \text{ mm}

is connected to a battery. The circuit has a resistance of

10 \text{ k}\Omega

. Determine the time constant of the circuit.

Options

A $17.7 \text{ ms}$
B $1.8 \mu\text{s}$
C $0.18 \mu\text{s}$
D $0.18 \text{ ms}$

Correct answer

C. $0.18 \mu\text{s}$

Step-by-step solution

The capacitance of a parallel-plate capacitor is given by

C = \dfrac{\varepsilon_0 A}{d}

Substituting the given values:

C = \dfrac{8.85 \times 10^{-12} \times 20 \times 10^{-4}}{1.0 \times 10^{-3}}

C = 17.7 \times 10^{-12} \text{ F}

The time constant of the RC circuit is given by

\tau = RC

\tau = 10 \times 10^3 \times 17.7 \times 10^{-12}

\tau = 17.7 \times 10^{-8} \text{ s} = 0.177 \mu\text{s} \approx 0.18 \mu\text{s}

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