Boundary Conditions on rrrr EHDB,,,

Chapter 9 General Boundary Conditions: Summary of Results

Vector form Scalar form Description rr r $ −= Tangential electric field, E , is 1. eEEn ¥-=di120 EEtt120 continuous. rr r $ −= 2. eHHjns¥-=di12 HHtts12 j The discontinuity of the tan- gentialH field equals the sur- face current. rr $ 3. eDDns◊ di12-=r DDnn12-=r sTher discontinuity of the normal D equals the surface charge density. rr r $ −= B 4. eBBn ◊ di12-=0 BBnn120 The normal component of is continuous.

Derivation of these most general forms is in the text, Chapter 4-9 and 5-7. Study them carefully. In what follows in these notes, we give the derivations of the same four boundary conditions for two important special cases. That is, (a) dielectric to perfect conductor boundary, and (b) general medium to general medium boundary with no surface charges and surface current den- sities. There are two ways to solve electromagnetic problems: 1. Find out charge and current distributions everywhere in space and solve Maxwell’s Equations everywhere. 2. Solve Maxwell’s Equations in a limited region of interest, subject to “boundary conditions” on the boundaries defining this region. Boundary condition means the value of the fields just at the boundary surface. The second method is used most often. It is especially useful when the boundaries are conductors. Boundary conditions at perfect conductors are fairly simple. We usually approximate good conductors as perfect conduc- tors in this class. Boundary Conditions at a Surface of Perfect Conductor Chapter 4-9 Text rr

Fields inside conductor are zero: EH22==00, . 1. Boundary condition for tangential component of electric field: Dielectric (perfect, or lossy) 1 a b s =• d c rr 2 Perfect Conductor EH22==00, r r r r r r r r r b c d a d r Edl◊◊◊◊+++=- Edl Edl Edl Bds ◊ zza b c zzz d dt abcd Area enclosed is infinitely small. Therefore RHS → 0 . b a “ad ” and “cb ” are infinitely small. Therefore, and Æ 0 . za zd T d E2 (inside conductor) is zero. Therefore, Æ 0 . zc This leaves: b r r Edl◊ = 0 za r If ab distance is finite, but small, then E is constant, and we have: r ⋅ =∆ = Et 0 Et component of E along ab; i.e., tangential component = Thus, Et 0 . In vector form, this can be written: r $ $ eEn ¥=0 , where en is unit normal. = A very important corollary of this boundary condition is that, since Et 0 at the surface of a perfect conductor,

b r r Edl◊ = 0 za between any point on or inside the conductor. Therefore, a perfect con- ductor is an equipotential surface.

2. Boundary condition for tangential component of magnetic field:

Text, 5-7

e$ d n a b ∆ is small, xxxxxxxxxxxxxxxxxxxxxxxxxxxx but finite d r c d Æ0 js r r r r d r r zzHdl◊◊=+z jds Dds ◊ abcddt abcd The second term on the right hand side goes to zero, since the area is infi- nitely small. However, the first term on the right hand side doesr not go to zero. The reason is that, although the area is rinfinitely small, j can be infinitely large.

This means that a surface current js can exist on the boundary, normal to the area abcd. We have already discussed this before. r r r U r dr Æ0 | jjs = ◊d V js finite j Æ•W| ∆ ⋅ Total current enclosed in the area abcd then is js . The direction of js → must be into the paper if Ht on ab is in the ab direction. Then, the left hand side becomes (by the same reasoning as in 1):

b r r Hdl◊ = Ht D za ∆∆= Thus, Hjts rr This also gives ∴=Hj, or eH$ ¥= j ts Anscorrect direction $ bgen is outward normal $ $ (en is frequently written more simply as n . This is the notation our text uses.)

3. Boundary condition on normal component of electric field:

small, but finite, area ∆A

U V| W| infinitesimal thickness d

r r z Dds◊ = z r dv surface of box volume of box

If there is only a volume charge density, the right hand side would be zero, since the volume of the box is infinitely small. However, there can be a surface charge density on the surface of the con- ductor. Because of that: right hand side = DA◊ rs r r r r r r left hand side = z Dds◊ + z Dds◊ + z Dds◊ side surface top bottom r r The integral of Dds◊ over the side surface is zero, since its height is infini- tesimal. The integral over the bottom surface is zero because the field is zero inside. r r r \=LHS Dds = DD A positive, if D z ◊ n is pointing away top from conductor

Thus, DDAADrsn= r $ Dns= r , or eDns◊ = r r ρ Note that s is positive for D pointing outward, as it should be (Field lines originate on positive charges).

4. Boundary condition on normal component of magnetic field: r r We apply z Bds◊ = 0 to the same pillbox as in 3. Then, obviously, since RHS = 0 , we get: r = $ Bn 0 , or eBn ◊ = 0

Summary of B.C. At the surface of a perfect conductor: r = $ = Et 0 eEn ¥=0 Tangential component of E 0. rr r = $ HjtseHjns¥= Tangential component of H = surface current density. r r $ Dns= r eDns◊ = r Normal component of D = surface charge density. r r = $ Bn 0 eBn ◊ = 0 Normal component of B = 0 .

Boundary conditions at boundary between two dielectrics (or two gen- r == eral media). No surface charges or currents: ss00, j We use exactly the same methods as we did in the previous sections. The main difference are that: a. There are fields on both sides of the boundary, since only inside per- fect conductors are the field’s zero. b. Under normal conditions, there is no surface charge on the boundary. c. There is no surface current on boundary, since surface currents can exist only on a perfect conductor surface. 1. Tangential Electric Field Component: ∆ r 1 a b Et1 r 2 d c Et2 b r r d r r Edl◊◊+= Edl 0 a zz c rr EEDD-=0 tt12rr rr $ \=EEtt12, or eEEn ¥-di12 =0

2. Tangential Magnetic Field Component: = Since js 0 , we get the same result as in 1. rr HH= $ tt12, or eHHn ¥-di12 =0

3. Normal Electric Field Component: r r r r Since rs = 0, z Dds◊ + z Dds◊ = 0 top bottom rr = $ DDnn12, or eDDn ◊di12-=0

4. Normal Magnetic Field Component: Same as 3. rr = $ BBnn12, or eBBn ◊di12-=0

5. Examples Useful for Remembering 2. and 3. H = 0 1 = ΟΟΟΟΟΟΟΟ Hjs parallel plate ×××××××× 2 H = 0 currents 3 1 E = 0 ++++++++ −−−−−−−− 2 E = res / parallel plate capacitor 3 E = 0 More on Boundary Conditions A. It can be easily shownr that an equivalent form of Boundary Condition 1. = (that is, tangential E continuous: EEtt12 ) is: = VV12 That is, potential is continuous at an interface. r $ B. Consider eDns◊ = r , which is the boundary condition for the normal component of the electric displacement at the interface between a per- fect conductor and a dielectric. rr r DE= e EV=-— ∂ V Then: EeV=-$ ◊ — =- nn ∂ n ∂V \=-re, s ∂n ∂V where means the normal derivative (that is, derivative in the direc- ∂n tion of the normal out of the conducting surface). Example shortly!

C. If the value of the potential V is given on a conducting surface, that too is called a boundary condition (to Laplace’s Equation).