The diagram shows a parallel-plate capacitor containing a slab of dielectric and

Question

The diagram shows a parallel-plate capacitor containing a slab of dielectric and an air gap. The following questions are concerned with the application of Gauss's law E0 integration kE.dS = q. where dS is the area The two cylinders, 3 and 4, have the same base area. The values of E0 integration kE.dS evaluated over the entire surfaces of cylinders 3 and 4 will be A. different because of the different values of dielectric constant at each of the upper surfaces; B. different because cylinder 3 includes only one layer of induced charge, while cylinder 4 includes both; C. the same because both cylinders contain the same free charge.

Explanation / Answer

Option C .Both the cylinders contain the same free charge

Also what would be the value of e0 integral of E dot dS evaluated over the WHOLE AREA of cylinder 2? 1) finite and negative because cylinder 2 contains the negative charge induced on the surface of the dielectric 2)finite and positive because k is larger at the lower surface of 2 then at the upper surface 3) zero because cylinder 2 contains no free charge

3) zero because cylinder 2 contains no free charge

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