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 37.4 pF and its working voltage is 500 kV.

Step by step solution

01

Convert units

Start by converting the given units into SI units. The area \(A\) should be in m², so we convert it from cm² to m². And the separation \(d\) should be in m, so we convert it from μm to m.
02

Calculate The Capacitance

To calculate the capacitance, we will use the formula \(C = \frac{\varepsilon_{0} \varepsilon_{r} A}{d}\). Where \(\varepsilon_{0} = 8.85 \times 10^{-12} F/m\), \(\varepsilon_{r} = 2.3\) for polyethylene, \(A\) is the area we obtained from part 1 and \(d\) is the separation we obtained from part 1.
03

Working Voltage

The breakdown field strength of polyethylene is \(E = 20 MV/m\), we will rearrange the formula \(E = \frac{V}{d}\) to solve for \(V\).
04

Calculate The Working Voltage

Substitute the given values into the formula to find the working voltage.

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