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Electrostatic Boundary Conditions

EE 141 Lecture NotesTopic 13

(5.3 of Text)

Professor K. E. OughstunSchool of Engineering

College of Engineering & Mathematical SciencesUniversity of Vermont

2009


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Motivation


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Dielectric Boundary Conditions


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Dielectric Boundary ConditionsLet the interface separating two dielectrics with permittivities1 & 2be denoted by S with surface normal n directed from medium 2 intomedium 1.

At any point r S the electric eld vectorE j (r ), j = 1 , 2, on eitherside of the interface S may be decomposed into tangential E tj (r ) and

normal E nj (r ) components with respect to the surface S at thatpoint as

E 1(r ) = E t 1(r ) + E n1(r ), (1)E 2(r ) = E t 2(r ) + E n2(r ), (2)

for all r S .

Notice that n changes as r S varies over the interface surfaceS and that this eld decomposition will then also vary as this directionchanges.


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Dielectric Boundary ConditionsApplication of the integral form of Faradays law to an innitesimallysmall loopC in the plane of the incident and transmitted electric eldvectors about the point r S , the upper side tangent to S inmedium 1 and the lower side tangent toS in medium 2, gives

C

E d =

b

a

E 2 d +

d

c

E 1 d = 0

where the contributions from the sides vanish as h 0 about S .In addition,

E 2 d = E t 2 , & E 1 d = E t 1

in the limit as 0 on S . Hence, in this limit, Faradays law givesE t 2 E t 1 = 0, or

E t 1(r ) = E t 2(r ), r S (3)

The tangential component of E is continuous across the interfaceS .


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Dielectric Boundary ConditionsApplication of the integral form of Gauss lawto an innitesimallysmall pillbox of thickness h 0 with upper surfaceS 1 parallel toS in medium 1 with outward normaln 1 = n and lower surfaceS 2parallel to S in medium 2 with outward normaln 2 = n , gives

S G D d s = S 1 D 1 n 1ds + S 2 D 2 n 2ds = s s where the contributions from the sides vanish as h 0 about S .Here s = s (r ), r S , denotes the surface charge density residingon the interface S . In the limit as s 0, one obtains

n D 1(r ) D 2(r ) = s (r ), r S (4)

orD n1(r ) D n2(r ) = s (r ), r S (5)

The normal component of D changes discontinuously across the

interface S by an amount given by the surface charge density s atthat point.


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Dielectric Boundary ConditionsAt the interface between two dielectrics with s = 0, the boundaryconditions are

1E n1 = 2E n2 ,E t 1 = E t 2 .

Let E 1 be at the angle 1 with respect to the surface normal n andE 2 be at the angle 2 with respect to the surface normal n , where

1 = arctanE t 1

E n1, 2 = arctan

E t 2

E n2.


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Dielectric Boundary ConditionsThen

tan

2 =

E t 2

E n2 =

2

1

E t 1

E n1 =

2

1 tan

1,

1 tan 1 = 2 tan 2 (6)

Notice that

1 > 2 = tan 1 > tan 2

2 > 1 = tan 2 > tan 1

d C di i h S f f


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Boundary Conditions at the Surface of aPerfect Conductor ( = )

A perfect conductor may be dened as a material inside whichelectric charge can freely ow.

In electrostatics one assumes that the charges have all reached theirequilibrium positions and are now xed in space. Hence, inside aconductor the electrostatic eld intensity E vanishes and all pointsare at the same potential; that is, a conductor forms an equipotential.

When a conductor is charged, the charges arrange themselves so that

the net electric eld due to all the charges is zero inside theconductor. If a conductor is placed in an electrostatic eld, thecharges temporarily ow within it in such a manner to produce asecond eld that, added to the rst, results in a net zero eld insidethe conductor. The eld outside the conductor is then distorted bythese charges, resulting in an altered static conguration.

B d C di i h S f f


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By Gauss lawfor an electrostatic eld, = E

= 0 within aconductor. Hence, any net static charge on a conductor must resideon its surface.

At the surface S of a conductor, the electrostatic eld intensity E (r )must be normal to S , for if it were not, there would be a tangentialcomponent of E that would cause the surface charge to ow alongthe surface. Hence, by Gauss law

E ext (r ) = s (r ) n , r S (7)

where n denotes the unit outward normal vector to S at the pointr S , s (r ) denotes the surface charge density at that point, andwhere E ext (r ) denotes the electrostatic eld just above the conductorsurface S at that point in a medium with dielectric permittivity.

B d C di i h S f f


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E ext (r )

E (r ) = 0 > 0

= 0

s (r )

Illustration of the external electrostatic eldE ext (r ) terminating on

the surface charge s (r) of a conductor surface S embedded in adielectric medium with permittivity.


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ProblemsProblem 15. Charge Q 1 is uniformly distributed over a sphericalsurface of radiusa surrounding a dielectric with permittivity1 , andcharge Q 2 is uniformly distributed over a spherical surface of radiusb where the dielectric permittivity is2 for a < r < b and 2 for r > b ,as illustrated. Apply Gauss law and the appropriate boundaryconditions to determine the electrostatic eld in each of the spherical

regions 0 r < a, a < r < b , and r > b .

Q 1

Q 2

1

2

3

boundry condition 1

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