Question: A parallel-plate capacitor is constructed using a dielectric whose constant varies with position. The plates have area A. The bottom plate is at y =

A parallel-plate capacitor is constructed using a dielectric whose constant varies with position. The plates have area A. The bottom plate is at y = 0 and the top plate is at y = y0. The dielectric constant is given as a function of y according to κ = 1 + (3/y0)y.

(a) What is the capacitance?

(b) Find σb/σf on the surfaces of the dielectric.

(c) Use Gauss's law to find the induced volume charge density ρ(y) within this dielectric.

(d) Integrate the expression for the volume charge density found in (c) over the dielectric, and show that the total induced bound charge, including that on the surfaces, is zero.

Step by Step Solution

3.50 Rating (163 Votes )

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock

a The expression for the electric field and integrate from y 0 to y y 0 E E 0 y sye 0 C sAV 3e 0 Ay ... View full answer

blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Document Format (1 attachment)

Word file Icon

10-P-E-C-D (407).docx

120 KBs Word File

Students Have Also Explored These Related Electricity and Magnetism Questions!