A parallel-plate capacitor with a plate area of 25.0 cm^2 and a 2.00 mm separation between its plates is connected to a 12.0 V battery. Determine the charge on the capacitor.

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Quantrex Academy · Accepted Answer

Correct answer: D. 1.33 × 10^ -10 C

The capacitance of a parallel-plate capacitor is given by

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

The charge on the capacitor is

Q = CV = \dfrac{\varepsilon_0 A V}{d}

Substituting the given values:

A = 25.0 \text{ cm}^2 = 25.0 \times 10^{-4} \text{ m}^2

d = 2.00 \text{ mm} = 2.00 \times 10^{-3} \text{ m}

V = 12.0 \text{ V}

\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m}

Q = \dfrac{8.85 \times 10^{-12} \times 25.0 \times 10^{-4} \times 12.0}{2.00 \times 10^{-3}}

Q = 8.85 \times 10^{-12} \times 25.0 \times 10^{-4} \times 6.0 \times 10^3

Q = 8.85 \times 10^{-12} \times 150 \times 10^{-1}

Q = 132.75 \times 10^{-12} \text{ C}

Q \approx 1.33 \times 10^{-10} \text{ C}

A parallel-plate capacitor with a plate area of 25.0 cm^2 and a...

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