Suppose the region abovethe xyplane in Ex. 4.8 is alsofilled withlinear dielectric but of a different susceptibility χ'e.Find the potential everywhere.

Short Answer

Expert verified

The potential is V=14πε0q/ε'rx2+y2+(zd)2+qbx2+y2+(z+d)2     z>014πε02q/(ε'r+εr)x2+y2+(zd)2                                     z<0

Step by step solution

01

Step 1:Given data

There is a point charge q.

There is a polarization charge surrounding q.

There is a surface chargeσb on the upper surface of the lower dielectric.

There is a surface charge σ'b on the lower surface of the upper dielectric.

The susceptibility of the medium is χe and χ'e.

02

Define the polarization charge

The polarization charge due to q is

qp=-qχ'e1+χ'e

03

Derive the expression for the potential

The expressions for the surface bound charge densities are

σb=ε0χe14πε0qdε'r(r2+d2)3/2σb2ε0σ'b2ε0σ'b=ε0χ'e14πε0qdε'r(r2+d2)3/2σb2ε0σ'b2ε0

Here, ε0is the permittivity of free space and ε'ris the permittivity of the upper medium.

Solve the above equations to get

σb=14πqd(r2+d2)3/2χe1+χe+χ'e2σ'b=14πqd(r2+d2)3/2εrχ'e/ε'r1+χe+χ'e2

Here, εris the permittivity of the lower medium.

The total bound surface charge is then

σ=14πqd(r2+d2)3/2χ'eχeε'r1+χe+χ'e2

The total bound charge from the surface charge density is

qb=qχ'eχe2ε'r1+χe+χ'e2

The potential is thus

V=14πε0q/ε'rx2+y2+(zd)2+qbx2+y2+(z+d)2     z>014πε02q/(ε'r+εr)x2+y2+(zd)2                                     z<0

Thus, this is the expression for the potential everywhere.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free