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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 with a plate area of 25 cm^2 and a...

A parallel-plate capacitor with a plate area of 25 cm225 \text{ cm}^2 and a separation of 1.00 mm1.00 \text{ mm} is connected to a 6.0 V6.0 \text{ V} battery. Determine the charge that has flown through the battery and the work done by the battery during this process.

Options

  1. A6.65×10−11 C6.65 \times 10^{-11} \text{ C}, 4.0×10−10 J4.0 \times 10^{-10} \text{ J}
  2. B2.66×10−10 C2.66 \times 10^{-10} \text{ C}, 1.6×10−9 J1.6 \times 10^{-9} \text{ J}
  3. C1.33×10−10 C1.33 \times 10^{-10} \text{ C}, 8.0×10−10 J8.0 \times 10^{-10} \text{ J}
  4. D1.33×10−10 C1.33 \times 10^{-10} \text{ C}, 4.0×10−10 J4.0 \times 10^{-10} \text{ J}

Correct answer

C. 1.33×10−10 C1.33 \times 10^{-10} \text{ C}, 8.0×10−10 J8.0 \times 10^{-10} \text{ J}

Step-by-step solution

The capacitance of the parallel-plate capacitor is given by:

C=ε0AdC = \dfrac{\varepsilon_0 A}{d}

Substituting the given values:

C=8.85×10−12×25×10−41.00×10−3=2.2125×10−11 FC = \dfrac{8.85 \times 10^{-12} \times 25 \times 10^{-4}}{1.00 \times 10^{-3}} = 2.2125 \times 10^{-11} \text{ F}

The charge that flows through the battery is:

Q=CV=2.2125×10−11×6.0=1.3275×10−10 C≈1.33×10−10 CQ = CV = 2.2125 \times 10^{-11} \times 6.0 = 1.3275 \times 10^{-10} \text{ C} \approx 1.33 \times 10^{-10} \text{ C}

The work done by the battery is the product of the charge that flows through it and its electromotive force (voltage):

W=QV=1.3275×10−10×6.0=7.965×10−10 J≈8.0×10−10 JW = QV = 1.3275 \times 10^{-10} \times 6.0 = 7.965 \times 10^{-10} \text{ J} \approx 8.0 \times 10^{-10} \text{ J}
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