A circular parallel plate capacitor with radius 6cm is separated by 0.12cm. Calculate the capacitance of the capacitor [\(\pi\) = 3.142, ε\(_0\) = 8.85 x 10\(^{-12}\)Nm\(^2\)C\(^2\)]
The capacitance \( C \) of a parallel-plate capacitor is calculated using the formula:
\(C = \epsilon_0 \frac{A}{d}\)
where: \(\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m},\)
\(A = \pi r^2 \quad \text{(area of one plate)},\)
and \( d \) is the separation between plates. \(r = 6 \, \text{cm} = 0.06 \, \text{m},\)
\(d = 0.12 \, \text{cm} = 0.0012 \, \text{m}\)
Calculate the area: \(A = \pi r^2 = 3.142 \times (0.06)^2 = 3.142 \times 0.0036 = 0.0113112 \, \text{m}^2\)
Calculate the capacitance: \(C = 8.85 \times 10^{-12} \times \frac{0.0113112}{0.0012} = 8.85 \times 10^{-12} \times 9426 \approx 8.34 \times 10^{-11} \, \text{F}\)
There is an explanation video available below.
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