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Chapter 23: Problem 49

A parallel-plate capacitor has plates with area \(50 \mathrm{cm}^{2}\) separated by \(25 \mu \mathrm{m}\) of polyethylene. Find its (a) capacitance and (b) working voltage.

Short Answer

Expert verified
The capacitance of the capacitor is calculated to be approx. 4.14 nF, and the working voltage (breakdown voltage) of the capacitor is 15 kV.

Step by step solution

01

Convert given values to SI units

Convert the given values for area and distance into SI units. \(1 m = 100 cm\), so \(50 \mathrm{cm^2} = 0.005 \mathrm{m^2}\), and \(1 \mu \mathrm{m} = 10^{-6} m\), so \(25 \mu \mathrm{m} = 2.5 \times 10^{-5} \mathrm{m}\).
02

Calculate Capacitance

Using the formula for capacitance, where \(蔚_0\)= \(8.854 \times 10^{-12}\) \(F/m\), \(蔚_r\) for polyethylene = 2.3, \(A = 0.005 m^2\), and \(d = 2.5 \times 10^{-5} m\), substitute these values into the formula to calculate the capacitance. \(C = \frac{8.854 \times 10^{-12} F/m \times 2.3 \times 0.005 m^2}{2.5 \times 10^{-5} m}\)
03

Calculate Working Voltage

Knowing that the breakdown voltage (maximum voltage it can handle) for polyethylene is approximately \(600 kV/mm\) and the plates are separated by \(25 \mu m = 0.025mm\), multiply these two values to get the working voltage. So the working voltage = \(0.025mm \times 600 kV/mm\)

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