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 =

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.

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

This is in agreement with the result of article 1.14

1. Continuous charge distribution: an arrangement of many discrete charges so closely spaced that the charge is treated as a continuum, resulting in a replacement of discrete sums with integrals.
2. Distribution of charges can be Line charge density, Surface charge density and Volume charge density
3. Electric field for a continuous charge distribution can be written as =-
4. Electric field of a line segment or rod is given by

5.     Electric field of an infinite line of charge

6.     Electric field due to a ring of charge

7.     Electric potential is the work done by an applied force on a unit charge bringing it from infinity to a specific point. The work done is called electric 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:

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.

In one dimension the electrostatic potential V depends on only one variable x. The electrostatic potential V(x) is a solution of the one-dimensional Laplace equation

=0

The general solution of this equation is

V(x) = sx + b

where s and b are arbitrary constants. These constants are fixed when the value of the potential is specified at two different positions.

Consider a one-dimensional world with two point conductors located at x = 0 m and at x = 10 m. The conductor at x = 0 m is grounded (V = 0 V) and the conductor at x = 10 m is kept at a constant potential of 200 V. Determine V.

The boundary conditions for V are

V(0) = b = 0V

and

V(10) = 10s +b =200V

The first boundary condition shows that b = 0 V. The second boundary condition shows that s = 20 V/m. The electrostatic potential for this system of conductors is thus

V(x) = 20x
The corresponding electric field can be obtained from the gradient of V

E(x) = = -20 V/m

The boundary conditions used here, can be used to specify the electrostatic potential between x = 0 m and x = 10 m but not in the region x < 0 m and x > 10 m.

If the solution obtained here was the general solution for all x, then V would approach infinity when x approaches infinity and V would approach minus infinity when x approaches minus infinity. The boundary conditions therefore provide the information necessary to uniquely define a solution to Laplace's equation, but they also define the boundary of the region where this solution is valid (in this example 0 m < x < 10 m).

The following properties are true for any solution of the one-dimensional Laplace equation:
Property 1:

 V(x) is the average of V(x + R) and V(x - R) for any R as long as x + R and x - R are located in the region between the boundary points. This property is easy to proof:

This property immediately suggests a powerful analytical method to determine the solution of Laplace's equation. If the boundary values of V are

V(x=a) =Va

and

V(x=b) =Vb

then property 1 can be used to determine the value of the potential at (a + b)/2:
 

Next we can determine the value of the potential at x = (3 a + b)/4 and at

x = (a + 3 b)/4 :

This process can be repeated and V can be calculated in this manner at any point between x = a and x = b (but not in the region x > b and x < a).

Property 2:

The solution of Laplace's equation cannot have local maxima or minima. Extreme values must occur at the end points (the boundaries). This is a direct consequence              of              property              1.
 

Property 2 has an important consequence: a charged particle cannot be held in stable equilibrium by electrostatic forces alone (Earnshaw's Theorem). A particle is in a stable equilibrium if it is located at a position where the potential has a minimum value.

A small displacement away from the equilibrium position will increase the electrostatic potential of the particle, and a restoring force will try to move the particle back to its equilibrium position.

There can be no local maxima or minima in the electrostatic potential.

The particle cannot be held in stable equilibrium by just electrostatic forces.

+ =0

Property 1:

The value of V at a point (x, y) is equal to the average value of V around this point

where the path integral is along a circle of arbitrary radius, centered at (x, y) and with              radius                            R.

 

Property 2:
V has no local maxima or minima; all extremes occur at the boundaries.

+ =0

Property 1:
The value of V at a point (x, y, z) is equal to the average value of V around this point

Where the surface integral is across the surface of a sphere of arbitrary radius, centered at (x,y,z) and with radius R.

Figure 8: Laplace's equation in three dimensions

Proof of property 1.

To proof this property of V consider the electrostatic potential generated by a point charge q located on the z axis, a distance r away from the center of a sphere of radius R . The potential at P, generated by charge q, is equal to

Vp=

where d is the distance between P and q. Using the cosine rule we can express d in terms of r, R and θ

= + -2Rrcosθ

The potential at P due to charge q is therefore equal to

Vp=

The average potential on the surface of the sphere can be obtained by integrating Vp across the surface of the sphere. The average potential is equal to

Which is equal to the potential due to q at the center of the sphere. Applying the principle of superposition it is easy to show that the average potential generated by a collection of point charges is equal to the net potential they produce at the center               of              the              sphere.

 

Property 2:
The electrostatic potential V has no local maxima or minima; all extremes occur at the              boundaries.
 

Find the general solution to Laplace's equation in spherical coordinates, for the case where V depends only on r. Then do the same for cylindrical coordinates.

Laplace's equation in spherical coordinates is given by

If V is only a function of r then

=0

and

=0

Therefore, Laplace's equation can be rewritten as

The solution V of this second-order differential equation must satisfy the following first-order differential equation:

This differential equation can be rewritten as

=
The general solution of this first-order differential equation is

V(r) = +b

Where b is a constant. If V = 0 at infinity then b must be equal to zero, and consequently

V(r)=
Laplace's equation in cylindrical coordinates is

If V is only a function of r then

=0

and

=0

Therefore, Laplace's equation can be rewritten as

The solution V of this second-order differential equation must satisfy the following first-order differential equation:

r= a = constant

This differential equation can be rewritten as

=

The general solution of this first-order differential equation is

V(r) = aln(r) +b

Where b is a constant. The constants a and b are determined by the boundary conditions.

1. Therefore, the potential is related to the charge density by Poisson's equation = 2V =
2. In a charge-free region of space, this becomes Laplace's equation 2V = 0
3. For 1D - V(x) is the average of V(x + R) and V(x - R) for any R as long as x + R and x - R are located in the region between the boundary points.
4. For 1D - The solution of Laplace's equation cannot have local maxima or minima. Extreme values must occur at the end points (the boundaries).
5. For 2D - The value of V at a point (x, y) is equal to the average value of V around this point
    
6. For 2D - V has no local maxima or minima; all extremes occur at the boundaries.
7. For 3D - The value of V at a point (x, y, z) is equal to the average value of V around this point

8.     For 3D - The electrostatic potential V has no local maxima or minima; all extremes occur at the              boundaries.
 

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

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.

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.

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.

1. It is a very powerful technique for solving electrostatics problems involving charges and conductors.
2. It turns out that an electrically equivalent problem can be created using the image distribution with the ground plane removed.
3. This technique of using image charges to obtain the electrostatic potential in some region of space is called the method of images.
   The electrostatic energy of the image-charge system is equal to Wimage =
4. The electrostatic energy of the real system is therefore equal to Wreal = Wimage =

= = A

- =

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.

1. It’s found that the electric field undergoes a discontinuity as you cross the boundary.
2. Understanding this boundary condition helps us to determine the electric field above some material with some surface charge.
3. It allows us to determine the electric field even if external electric fields are present.
4. 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  - =
5. The tangential component is always continuous and therefore contributes nothing to the discontinuity of the total electric field.
6. As the distance of the path shrinks to zero, where we just cross the surface, then:

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  .

Case (ii) Electric field due to an electric dipole at a point on the equatorial plane

Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane as shown in Figure 17.

Since the point C is equidistant from +q and –q, the magnitude of the electric fields of +q and –q are the same. The direction of  is along BC and the direction of  is along CA.  and  are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it. The perpendicular components |  | sinθ and | | sinθ are oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the parallel components of + and  and its direction is along - as shown in the Figure 17.

    ………(8)

Figure 17: Electric field of the dipole at a point on the equatorial plane

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.

1. Electric field due to an electric dipole at points on the axial line is given as

2.     Electric field due to an electric dipole at a point on the equatorial plane At very large distances (r>>a) given as

3.     Potential due to an electric dipole given as

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.

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

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.