A parallel plate capacitor has a paper dielectric having dielectric strength 8.0 kV/mm and dielectric constant 3.0. The plate area is 3000 cm² and the plate separation is 0.50 mm. What is the capacitance?

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Quizplus · Accepted Answer

The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\varepsilon_0 \varepsilon_r A}{d}$, where $\varepsilon_0$ is the vacuum permittivity ($8.85 \times 10^{-12} \, \text{F/m}$), $\varepsilon_r$ is the relative permittivity (dielectric constant), $A$ is the area of the plates in square meters, and $d$ is the separation between the plates in meters. Given $\varepsilon_r = 3.0$, $A = 3000 \, \text{cm}^2 = 3000 \times 10^{-4} \, \text{m}^2$, and $d = 0.50 \, \text{mm} = 0.50 \times 10^{-3} \, \text{m}$, the capacitance is calculated as follows:$$C = \frac{(8.85 \times 10^{-12} \, \text{F/m}) \times 3.0 \times (3000 \times 10^{-4} \, \text{m}^2)}{0.50 \times 10^{-3} \, \text{m}} = \frac{(8.85 \times 10^{-12}) \times 3 \times 0.3}{0.5 \times 10^{-3}}$$$$C = \frac{26.55 \times 10^{-12}}{0.5 \times 10^{-3}} = 53.1 \times 10^{-9} \, \text{F} = 53.1 \, \text{nF}$$However, given the options, it seems there was a calculation mistake in the final step. Correctly calculating gives:$$C = \frac{8.85 \times 3 \times 0.3}{0.5} \times 10^{-9} = 15.93 \times 10^{-9} \, \text{F} = 15.93 \, \text{nF}$$Rounding to the nearest option gives 16 nF, which is option A.

A Parallel Plate Capacitor Has a Paper Dielectric Having Dielectric