A parallel plate capacitor initially has a capacitance C₁and consists of two plates separated by a vacuum with a plate separation of dinit. The capacitor is then modified by introducing two different dielectric materials between the plates. The first dielectric material has a thickness of d₁and a relative permittivity of Er1, while the second material has a thickness of d₂ and a relative permittivity of r2. The plate separation is adjusted to accommodate the dielectric material without altering the plate area. After the introduction of the dielectrics, the new capacitance C₂ is measured to be half the initial capacitance C₁. Given that the thickness of the first dielectric is twice that of the second, and the sum of the thicknesses is equal to the original plate separation, demonstrate that the relative permittivity &r2 can be formulated as Er2 = Er1 6Er1 - 2

A parallel plate capacitor initially has a capacitance C₁and consists of two plates separated by a vacuum with a plate separation of dinit. The capacitor is then modified by introducing two different dielectric materials between the plates. The first dielectric material has a thickness of d₁and a relative permittivity of Er1, while the second material has a thickness of d₂ and a relative permittivity of r2. The plate separation is adjusted to accommodate the dielectric material without altering the plate area. After the introduction of the dielectrics, the new capacitance C₂ is measured to be half the initial capacitance C₁. Given that the thickness of the first dielectric is twice that of the second, and the sum of the thicknesses is equal to the original plate separation, demonstrate that the relative permittivity &r2 can be formulated as Er2 = Er1 6Er1 - 2

Delmar's Standard Textbook Of Electricity

7th Edition

ISBN:9781337900348

Author:Stephen L. Herman

Publisher:Stephen L. Herman

Chapter19: Capacitors

Section: Chapter Questions

Problem 3RQ: A capacitor uses air as a dielectric and has a capacitance of 3 F. A dielectric material is inserted...

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