Chapter 6: Problem 3
Capacitors tend to be more expensive as their capacitance and maximum voltage \(V_{\max }\) increase. The voltage \(V_{\max }\) is limited by the field strength at which the dielectric breaks down, \(E_{B D}\). Which of these dielectrics will give the largest \(C V_{\max }\) product for equal plate areas? \((a)\) Air: \(\epsilon_{r}=1\), \(E_{B D}=3 \mathrm{MV} / \mathrm{m} .(b)\) Barium titanate: \(\epsilon_{r}=1200, E_{B D}=3 \mathrm{MV} / \mathrm{m} .(c)\) Silicon dioxide: \(\epsilon_{r}=3.78, E_{B D}=16 \mathrm{MV} / \mathrm{m} .(d)\) Polyethylene: \(\epsilon_{r}=2.26, E_{B D}=\) \(4.7 \mathrm{MV} / \mathrm{m} .\)
Short Answer
Step by step solution
Key Concepts
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