Consider the charge distribution shown in Figure. Taking infinity as our reference point with zero potential, the electric potential at P due to dq is

Summing over contributions from all differential elements, we have

We established the relation between and V. If we consider two points which are separated by a small distance , the following differential form is obtained.

dV = -

In Cartesian coordinates,

E = Ex + Ey + Ez

and = dx + dy + dz

dV = ( Ex + Ey + Ez) .( dx + dy + dz)

dV = ( Ex dx+ Eydy + Ezdz

This implies

Ex =   Ey =   Ez =  

By introducing a differential quantity called the “del (gradient) operator”

The electric field can be written as

=-

Notice that ∇ operates on a scalar quantity (electric potential) and results in a vector quantity (electric field). Mathematically, we can think of as the negative of the gradient of the electric potential V. Physically, the negative sign implies that if V increases as a positive charge moves along some direction, say x, with , then there is a non-vanishing component of in the opposite direction(-Ex )

In the case of gravity, if the gravitational potential increases when a mass is lifted a distance h, the gravitational force must be downward.

If the charge distribution possesses spherical symmetry, then the resulting electric field is a function of the radial distance r, i.e. = Er

In this case, .dV=- Er dr.

 If V(r) is known, then may be obtained as

For example, the electric potential due to a point charge q is

V =

Using the above formula, the electric field is simply

E =

A useful approach to the calculation of electric potentials is to relate that potential to the charge density which gives rise to it. The electric field is related to the charge density by the divergence relationship

=

E = Electric field

= Charge density

= Permittivity

and the electric field is related to the electric potential by a gradient relationship

E =

Therefore the potential is related to the charge density by Poisson's equation

= 2V =

In a charge-free region of space, this becomes Laplace's equation

2V = 0

This mathematical operation, the divergence of the gradient of a function, is called the Laplacian. Expressing the Laplacian in different coordinate systems to take advantage of the symmetry of a charge distribution helps in the solution for the electric potential V.

For example, if the charge distribution has spherical symmetry, you use the Laplacian in spherical polar coordinates.

Since the potential is a scalar function, this approach has advantages over trying to calculate the electric field directly. Once the potential has been calculated, the electric field can be computed by taking the gradient of the potential.

The flux of electric field lines through any surface is proportional to the number of field lines passing through that surface. Consider for example a point charge q located at the origin. The electric flux E through a sphere of radius r, centered on the origin, is equal to

E = surface  

Since the number of field lines generated by the charge q depends only on the magnitude of the charge, any arbitrarily shaped surface that encloses q will intercept the same number of field lines. Therefore the electric flux through any surface that encloses the charge q is equal to q/0.Using the principle of superposition we can extend our conclusion easily to systems containing more than one point charge:

E = surface   = =

We thus conclude that for an arbitrary surface and arbitrary charge distribution

E = surface  =

where Qenclosed is the total charge enclosed by the surface. This is called Gauss's law. Since this equation involves an integral it is also called Gauss's law in integral form. Using the divergence theorem the electric flux E can be rewritten as

E = surface  = d

We can also rewrite the enclosed charge Qencl in terms of the charge density ρ

=d

Gauss's law can thus be rewritten as

d = d

Since we have not made any assumptions about the integration volume this equation must hold for any volume. This requires that the integrands are equal:

=

This equation is called Gauss's law in differential form.

This is the required expression for divergence of electric field.

Divergence is the outflow of flux from a small closed surface area (per unit volume) as volume shrinks to zero.

For a point charge q placing at the origin, the electric field is:

E =

The curl calculation by integration:

By superposition rule:

= ++++…………

= + +………=0

The curl calculation by differentiation:

Physical Interpretation of the Curlƒ

Consider a vector field F that represents a fluid velocity:

The curl of F at a point in a fluid is a measure of the rotation of the fluid.

If there is no rotation of fluid anywhere then ∇x F= 0. Such a vector field is said to be irrotational or conservative.

An electrostatic field (denoted by E) has the property ∇xE= 0, an irrotational (conservative) field.

Since it is a finite line segment, from far away, it should look like a point charge. We will check the expression we get to see if it meets this expectation.

The electric field for a line charge is given by the general expression

 =        …………..(1)

The symmetry of the situation (our choice of the two identical differential pieces of charge) implies the horizontal (x)-components of the field cancel, so that the net field points in the -direction. Let’s check this formally.

The total field  is the vector sum of the fields from each of the two charge elements are E1 and E2.

 = E1+ E2 = + + +   …………..(2)

Because the two charge elements are identical and are the same distance away from the point P where we want to calculate the field, =  , so those components cancel. This leaves

 = E1+ E2 = + = E1Cosθ+ E2Cosθ  …………..(3)

These components are also equal, so we have

…………..(4)

Here our differential line element dl is dx, in this example, since we are integrating along a line of charge that lies on the x-axis.

The limits of integration are 0 to  , not  to L, because we have constructed the net field from two differential pieces of charge dq. If we integrated along the entire length, we would pick up an erroneous factor of 2.

In principle, this is complete. However, to actually calculate this integral, we need to eliminate all the variables that are not given. In this case, both r  and θ change as we integrate outward to the end of the line charge, so those are the variables to get rid of as follow

r =

and

Cos θ = =         …………..(5)

Substituting, we obtain

          …………..(6)

This simplifies to

    …………..(7)

With the use of symmetry we are able to find electric field due to line element. This is a very common strategy for calculating electric fields. The fields of non-symmetrical charge distributions have to be handled with multiple integrals.

Electric field of an infinite line of charge

Find the electric field a distance z above the midpoint of an infinite line of charge that carries a uniform line charge density λ.

This is exactly like the 1.12 article except the limits of integration will be - to +  .

Again, the horizontal components cancel out, so we wind up with

Where our differential line element dl is dx, in this example, since we are integrating along a line of charge that lies on the x-axis. Again,

Cos θ = =

Substituting, we obtain

This simplifies to

 In the case of a finite line of charge, note that for z >>L, z2 dominates L the  in the denominator, so that Equation simplifies to

If you recall that λL=q, the total charge on the wire, we have retrieved the expression for the field of a point charge, as expected.

In the limit L, on the other hand, we get the field of an infinite straight wire, which is a straight wire whose length is much, much greater than either of its other dimensions, and also much, much greater than the distance at which the field is to be calculated:

 =        …………..(1)

A general element of the arc between θ and dθ is of length Rdθ and therefore contains a charge equal to λRdθ. The element is at a distance of r =  from P, the angle is ,

Cos = =

And therefore the electric field is

Significance

As usual, symmetry simplified this problem, in this particular case resulting in a trivial integral. Also, when we take the limit of Z >> R, we find that

Consider a differential element of length dx′ which carries a charge dq =λdx′, as shown in Figure. The source element is located at (x′,0) while the field point P is located on the y-axis at (0,y). The distance from dx′ to P is r = . Its contribution to the potential is given by

dV =

Taking V to be zero at infinity, the total potential due to the entire rod is

Where we have used the integration formula

In the limit the potential becomes l>>y

The corresponding electric field can be obtained as

Consider a small differential element dl= Rd on the ring. The element carries a charge

dq =λdl = λ        ………….(1)

and its contribution to the electric potential at P is

The electric potential at P due to the entire ring is

         ………….(2)

Where we have substituted Q=2πRλ  for the total charge on the ring. In the limit z>>R, the potential approaches its “point-charge” limit:

From Equation 2, the z-component of the electric field may be obtained as

To proof the first uniqueness theorem we will consider what happens when there are two solutions V1 and V2 of Laplace's equation in the volume shown in Figure. Since V1 and V2 are solutions of Laplace's equation we know that

=0

=0

Since both V1 and V2 are solutions, they must have the same value on the boundary. Thus V1 = V2 on the boundary of the volume. Now consider a third function V3, which is the difference between V1 and V2

=-

The function V3 is also a solution of Laplace's equation. This can be demonstrated easily:

=- =0

The value of the function V3 is equal to zero on the boundary of the volume since V1 = V2 there. However, property 2 of any solution of Laplace's equation states that it can have no local maxima or minima and that the extreme values of the solution must occur at the boundaries. Since V3 is a solution of Laplace's equation and its value is zero everywhere on the boundary of the volume, the maximum and minimum value of V3 must be equal to zero. Therefore, V3 must be equal to zero everywhere. This immediately implies that

V1 = V2

 This proves that there can be no two different functions V1 and V2 that are solutions of Laplace's equation and satisfy the same boundary conditions. Therefore, the solution of Laplace's equation is uniquely determined if its value is a specified function on all boundaries of the region. This also indicates that it does not matter how you come by your solution: if (a) it is a solution of Laplace's equation, and (b) it has the correct value on the boundaries, then it is the right and only solution.

Figure 10: Boundary

The first uniqueness theorem can only be applied in those regions that are free of charge and surrounded by a boundary with a known potential (not necessarily constant). In the laboratory the boundaries are usually conductors connected to batteries to keep them at a fixed potential.

 In many other electrostatic problems we do not know the potential at the boundaries of the system. Instead we might know the total charge on the various conductors that make up the system (note: knowing the total charge on a conductor does not imply a knowledge of the charge distribution ρ since it is influenced by the presence of the other conductors).

In addition to the conductors that make up the system, there might be a charge distribution ρ filling the regions between the conductors. For this type of system the first uniqueness theorem does not apply.

The second uniqueness theorem states that the electric field is uniquely determined if the total charge on each conductor is given and the charge distribution in the regions between the conductors is known.
 

The proof of the second uniqueness theorem is similar to the proof of the first uniqueness theorem.

Suppose that there are two fields and that are solutions of Poisson's equation in the region between the conductors. Thus

.=

.=

where ρ is the charge density at the point where the electric field is evaluated. The surface integrals of and , evaluated using a surface that is just outside one of the conductors with charge Qi, are equal to . Thus

=

=

The difference betweenand  , So

= -

 satisfies the following equations:

.= . - .  -   =0

=   - =  -  =0

Consider the surface integral of , integrated over all surfaces (the surface of all conductors and the outer surface). Since the potential on the surface of any conductor is constant, the electrostatic potential associated with and  , must also be constant on the surface of each conductor.

Therefore, =- will also be constant on the surface of each conductor.

The surface integral of over the surface of conductor i can be written as

= =0

Since the surface integral of over the surface of conductor i is equal to zero,

the surface integral of over all conductor surfaces will also be equal to zero. The surface integral of over the outer surface will also be equal to zero since on this surface. Thus

=0

The surface integral of can be rewritten using Green's identity as

0 = = - =

= - = - =0

where the volume integration is over all space between the conductors and the outer surface. Since is always positive, the volume integral of can only be equal to zero if everywhere. This implies immediately that , everywhere, and proves the second uniqueness theorem.

In solving some electrostatic problems dealing with some surface charge of density , it’s found that the electric field undergoes a discontinuity as you cross the boundary. Understanding this boundary condition helps us to determine the electric field above some material with some surface charge. It allows us to determine the electric field even if external electric fields are present. The charge density could even vary on the surface. Now, what is the boundary condition? That is, by what amount is the electric field discontinuous across a charged surface?

We will use Gauss’s law which is:

= = A

Now we’ll take a surface of charge density and surround a portion of the surface with a Gaussian “pillbox” of width and area A as shown in figure 13. Notice that it extends above and below the surface. Notice that both and point up.

This is to say that upward is the positive direction and to account for the fact that we don’t know what the direction of  E actually is. After all, as I said before, there could be an external field, and differences in charge density on the surface could alter the electric field in other areas. This also highlights something important about Gauss’s law. It means that E is not simply the field due to .

In general when we solve a problem and apply Gauss’s law, we are enclosing all the charge of the system to determine the electric field, and so E is due to the enclosed charge. But in this case, the electric field could be influenced by some outside source or other areas on the surface. What matters is that any outside electric fields contribute nothing to the flux.

Figure 13: Boundary Conditions

 In the limit that goes to zero, the sides of the pillbox contribute nothing to the flux. Because the flux is the contributions of , the flux gives us the electric field perpendicular to our Gaussian surface.

Most applications of Gauss’s law take advantage of symmetry so that the electric field is parallel to da and so that E is constant over the Gaussian surface.

In this case, we still want E to be constant across our Gaussian surface but E may not be parallel to da. So we must make our surface area small enough so that the field is constant. Since the perpendicular component of E is parallel to da, and we’ve taken an area small enough so that E is constant, we can take E outside the integral. This gives:

- =

The minus sign comes from the fact that below the surface, da points downward. This means that as we cross the surface charge, the perpendicular component of the electric field will always have a discontinuity of .

Tangential component of the electric field

 We can determine the tangential component by exploiting the path independence property of the electric field. We can make a path tangential to the surface on top that comes back around tangentially to the bottom as shown in figure 14.

Figure 14: Tangential component of the electric field

Now we use

=0

The ends of the path have length   but as we make that go to zero, the ends contribute nothing to the integral. This gives:

  - = 0

Again, the minus sign comes from the fact that dl points in different directions on the top and bottom. This means that:

So the tangential component is always continuous and therefore contributes nothing to the discontinuity of the total electric field. Combining the components then gives:

=

Where is the unit vector that is perpendicular to the surface, pointing from below to above. The electric field is related to the electric potential, so what happens to the potential as we cross the surface?

If we have two points, a below the surface and b below the surface, then the potential difference between the two points is:

= -

As the distance of the path shrinks to zero, where we just cross the surface, then:

So the potential is continuous as we cross the surface. Since E =, the gradient of the potential inherits the discontinuity of E. This means:

=

Or

- = -

Where = is the normal derivative of the potential which is the rate of change in the direction perpendicular to the surface.

Consider a point charge q held as a distance d above an infinite grounded conducting plane as shown in Figure The electrostatic potential of this system must satisfy the following two boundary conditions:

V(x,y,z) =0

A direct calculation of the electrostatic potential cannot be carried out since the charge distribution on the grounded conductor is unknown.

Note: the charge distribution on the surface of a grounded conductor does not need to be zero.

Figure 12: Method of images

Consider a second system, consisting of two point charges with charges +q and -q, located at z = d and z = -d, respectively as shown in figure. The electrostatic potential generated by these two charges can be calculated directly at any point in space. At a point P = (x, y, 0) on the xy plane the electrostatic potential is equal to
 

Since this solution satisfies the boundary conditions, it must be the correct solution in the region z > 0 for the system shown in Figure. This technique of using image charges to obtain the electrostatic potential in some region of space is called the method of images.

The electrostatic potential can be used to calculate the charge distribution on the grounded conductor. Since the electric field inside the conductor is equal to zero, the boundary condition for shows that the electric field right outside the conductor is equal to

= =

Where σ is the surface charge density and is the unit vector normal to the surface of the conductor. Expressing the electric field in terms of the electrostatic potential V we can rewrite this equation as

Substituting the solution for V in this equation we find

Only in the last step of this calculation have we substituted z = 0. The induced charge distribution is negative and the charge density is greatest at (x = 0, y = 0, z = 0). The total charge on the conductor can be calculated by surface integrating of σ:

where. Substituting the expression for σ in the integral we obtain

As a result of the induced surface charge on the conductor, the point charge q will be attracted towards the conductor. Since the electrostatic potential generated by the charge image-charge system is the same as the charge-conductor system in the region where z > 0, the associated electric field (and consequently the force on point charge q) will also be the same. The force exerted on point charge q can be obtained immediately by calculating the force exerted on the point charge by the image charge. This force is equal to

There is however one important difference between the image-charge system and the real system. This difference is the total electrostatic energy of the system. The electric field in the image-charge system is present everywhere, and the magnitude of the electric field at (x, y, z) will be the same as the magnitude of the electric field at (x, y, -z).

On the other hand, in the real system the electric field will only be non-zero in the region with z > 0. Since the electrostatic energy of a system is proportional to the volume integral of the electrostatic energy of the real system will be 1/2 of the electrostatic energy of the image-charge system (only 1/2 of the total volume has a non-zero electric field in the real system). The electrostatic energy of the image-charge system is equal to

Wimage =

The electrostatic energy of the real system is therefore equal to

Wreal = Wimage =

The electrostatic energy of the real system can also be obtained by calculating the work required to be done to assemble the system. In order to move the charge q to its final position we will have to exert a force opposite to the force exerted on it by the grounded conductor. The work done to move the charge from infinity along the z axis to z = d is equal to
 

Which is identical to the result obtained using the electrostatic potential energy of the image-charge system.

For each of the six faces of the cube, there is a surface charge ρps. For the face located at x = L/2,

The total bound surface charge is

The bound volume charge density is given by

ρpv = .P = -(a+a+a)=-3a

and the total bound volume charge is

Qv = dv =-3a =-3aL3

Hence the total charge is

Qr= QS + Qv = 3aL3 -3aL3 =0

Electric field due to an electric dipole at points on the axial line 

Consider an electric dipole placed on the x-axis as shown in Figure 15. A point C is located at a distance of r from the midpoint O of the dipole along the axial line.

Figure 15: Electric field of the dipole on the axial line

The electric field at a point C due to +q is

    ………(1)

Since the electric dipole moment vector  is from –q to +q and is directed along BC, the above equation is rewritten as

     ………(2)

where p ^ is the electric dipole moment unit vector from –q to +q.

The electric field at a point C due to –q is

    ………(3)

Since +q is located closer to the point C than –q,  ,  is stronger than , Therefore, the length of the  vector is drawn larger than that of vector.

The total electric field at point C is calculated using the superposition principle of the electric field.

   ………(4)

    ………(5)

     ………(6)

Note that the total electric field is along , since +q is closer to C than –q.

The direction of , is shown in Figure 16.

Figure 16: Total electric field of the dipole on the axial line

If the point C is very far away from the dipole then (r >> a). Under this limit the term ( r2 − a2 )2 ≈ r4 . Substituting this into equation (6), we get

    ………(7)

If the point C is chosen on the left side of the dipole, the total electric field is still in the direction of  .

    ………(8)

The magnitudes + and  are the same and are given by

     ………(9)

By substituting equation (9) into equation (8), we get

     ………(10)

At very large distances (r>>a), the equation (10) becomes

    ………(11)

Since electric potential obeys superposition principle so potential due to electric dipole as a whole would be sum of potential due to both the charges +q and -q. Thus

       ………….(1)

where r1 and r2 respectively are distance of charge +q and -q from point R.

Now draw line PC perpendicular to RO and line QD perpendicular to RO as shown in figure. From triangle POC
cosθ=OC/OP = OC/a
therefore OC=acosθ similarly OD=acosθ
Now,
r1 = QR≅RD = OR-OD = r-acosθ
r2 = PR≅RC = OR+OC = r+acosθ


since magnitude of dipole is
|p| = 2qa

                    …………(2)

If we consider the case where r>>a then

        …………(3)

Again since pcosθ= p· where,    is the unit vector along the vector OR then electric potential of dipole is

           …………(4)
for r>>a

From above equation we can see that potential due to electric dipole is inversely proportional to r2 not 1/r which is the case for potential due to single charge.

Potential due to electric dipole does not only depend on r but also depends on angle between position vector r and dipole moment p.

Now let’s turn from pictures to math and calculate the net surface density of the bound charges in terms of the polarization

P = net dipole moment /volume     …………(1)

In general, the polarization is not uniform but varies on the macroscopic scale, maybe because the dielectric is non-uniform, maybe because it’s subject to a non-uniform electric field, maybe both.

In any case, mathematically the non-uniform P(r) acts as a macroscopic field. So let’s calculate the electric potential V(r) due to a general polarization field, and then reinterpret the result in terms of the bound charges.

Let’s start with the potential of a single ideal dipole,

    …………(2)

Where the second equality is just the mathematical identity, ∇(−1/r) =/r2. More generally, for a dipole located at some pointr0, the potential is

      …………(3)

Where the subscript r of ∇r indicates that the gradient is taken with respect to the 3 coordinates of the r vector rather than of the r0.

Eq. (3) easily generalizes to the potential of several dipole moments, and hence to the potential of a continuous density of the dipole moment, thus

    …………(4)

Note that the gradient here is taken with respect to the r vector — the point where we measure the potential — rather than the with respect to the integration variable r′= (x′, y′, z′)

However, since 1/|r−r′| depends only on the difference between the two position vectors, we may trade the gradient w.r.t.  r for the (minus) gradient w.r.t. r′ using

      …………(5)

Thus

   …………(6)

At this point, the gradient acts on the integration variable, so we may integrate by parts using

 …………(7)

and the Gauss theorem

        …………(8)

Note that both terms on the bottom line here look like Coulomb potentials of continuous charges. Specifically, the second term looks like the Coulomb potential of the volume charge density.

ρb(r′) =−∇ ·P(r′)        …………(9)

While the first term looks like the Coulomb potential of the surface charge density

σb(r′) =P(r′)·n        …………(10)

Where n is the unit vector normal to the dielectric’s surface at the point r′, hence P·d2A= (P·n)d2A=σbd2A.                                                                                                                …………(11)

Indeed, in terms of the ρb(r) and σb(r), eq. (8) becomes

…………(12)

Physically, we identify the σb=P·n as the net surface density of the bound charges and the ρb=−∇ ·P as the net volume density of the bound charges. Note: for general non-uniform polarizations P(r), the positive and the negative bound charge densities may mis-cancel not only on the surface of a dielectric but also inside its volume. However, for the uniform polarization there are no net volume bound charges but only the surface bound charges.

Besides the bound charges due to polarization, a dielectric material may also contain some extra charges which just happen to be there. For a lack of a better term, such extra charges are called free charges, just to contrast them from the bound charged due to polarization. Anyway, the net macroscopic electric field does not care for the origin of the electric charges or how we call them but only

net(r) =free(r) + bound(r);  net(r) = free(r)  + bound(r);    ………(1)

In particular, the Gauss Law in differential form says

0.E(r) = net(r) + net (coordinate ⟂ surface) + more function terms due to linear and point charges                                                                                                   ………(2)

For simplicity, let's ignore for a moment the surface, line, and point charges and focus on just the volume charge density. This gives us

0.E(r) = net(r) = free(r) + bound(r) = free(r) −∇ ·P

which we may rewrite as

∇.(ε0E+P) = free                 ………(3)

In light of this formula, the combination

D(r) =ε0E(r) +P(r)                ………(4)

called the electric displacement field obeys the Gauss Law involving only the free charges but not the bound charges,

∇.D(r) = free

Outside the dielectric there is no polarization, so we simply set

D(r) =ε0E(r)         ……(1)

Consequently, at the dielectric’s boundary, the D field has two sources of discontinuity:

 (1) The abrupt disappearance of the P term, and

(2) Discontinuity of the E field due to the net surface charge density.

If we focus on the component of the D vector ⊥ to the boundary, we get

disc (D⊥) =−Pinside⊥+ ε0disc(E⊥) =−P·n + σnet=−σbound + σnet=σfree                 ……(2)

Again, only the free surface charge affects this discontinuity, the bound surface charge cancels out between the P and the ε0E terms. In terms of the D field's divergence, the discontinuity of ∇.(ε0E+P) = free  becomes

()singular = disc (D⊥)(coordinate ⟂ surface)  = σfree(coordinate ⟂ surface)                                                                                                                                               ……(3)

Similar -function terms would obtain at free surface charges stuck inside the dielectric or outside it, and if we also allow for line or point free charges, there would also be (2) and (3)on the RHS. Altogether, for a most general geometry of free charges the complete Gauss Law for the D field becomes

∇.D(r) = free(r) + σfree(coordinate ⟂ surface)  + more -function terms due to free linear and point charges,

but only the free charges appear on the RHS here. All contributions of the bound charges cancel against the ∇.P term hiding inside the ∇.E on the LHS. Therefore, the integral form of the Gauss Law for the D field is

=Qfree [surrounded by S]     ……(4)

where the RHS includes all possible configurations of free charges surrounded by the surface S, but only the free charges. By the way, the surface S in this formula can be any complete surface | it can lie completely inside the dielectric, or completely outside it, or even cross the dielectric's surface | but the Gauss Law will work in any case, as long as we properly account for the net free charge enclosed within S.

For highly symmetric cases, we may calculate the D field just from the Gauss Law; more general geometries do not allow such shortcuts.

In fact, in general D(r) (anything) as in case of E(r) = (r)  ……(5)

Indeed, while the electrostatic tension field (Electric Field) always obeys ∇E = 0, for the displacement field we have

∇D = ε0∇ E + ∇P  which does not need to vanish  ……(6)

This issue becomes particularly acute at the dielectric's boundary. We know that despite the surface charge density of the bound charges, the potential and the tangential components of the electric tension field must be continuous across the boundary.

V(just outside) =V(just inside)       ……(7)

E∥(just outside) =E∥(just inside)       ……(8)

On the other hand, the polarization field P abruptly drops to zero at the boundary, so if P just inside the dielectric is directed at some general angle to the surface, then both its tangential and normal components suffer discontinuities.

P∥(just inside) P∥(outside) = 0       ……(9)

P⟂ (just inside) P⟂(outside) = 0       ……(10)

Figure 22: Boundary Conditions On Displacement

Combining equations (9) and (10), we find that in general the tangential components of the displacement field are discontinuous across the dielectric's boundary,

D∥(just outside) D∥(just inside)       ……(11)

As to the normal components of the tension and the displacement fields, the E⟂ is generally discontinuous across the boundary due to surface bound charges, but as we saw in eq. (2),

In the absence of free charges

D⟂ (just outside) =D⟂ (just inside)       ……….(12)

In the middle of a dielectric or outside it:

∇.D=free        ……….(13)

∇E= 0        ………(14)

On the other hand,

∇.E= (1/ε0)net    which is often unknown    ………(15)

∇D= unknown        ………(16)

At the outer boundary of a dielectric, or at the boundary between two different dielectrics

• V, E∥, and D⟂ are continuous
•  E⟂ and D∥ are discontinuous

To complete this equation system, we need a relation between the E and the D fields at the same point r, and that relation depends very much on the dielectric material in question. For some material, such relation can be rather complicated, or even history-dependent. Fortunately, for most common dielectrics the relation between the E and the D fields is linear (unless the fields become too strong). So in this class, we shall hence forth focus on such linear dielectrics.

a) We apply Coulomb's or Gauss's law to obtain

(b) Using Coulomb's law, we have