Electric displacement field

The electric displacement field "D" is defined as\mathbf{D} \equiv \varepsilon_{0} \mathbf{E} + \mathbf{P},where \varepsilon_{0} is the vacuum permittivity (also called permittivity of free space), E is the electric field, and P is the (macroscopic) density of the permanent and induced electric dipole moments in the material, called the polarization density. The displacement field satisfies Gauss's law in a dielectric: \nabla\cdot\mathbf{D} = \rho -\rho_\text{b} = \rho_\text{f} In this equation, \rho_\text{f} is the number of free charges per unit volume. These charges are the ones that have made the volume non-neutral, and they are sometimes referred to as the space charge. This equation says, in effect, that the flux lines of D must begin and end on the free charges. In contrast \rho_\text{b}, which is called the bound charge, is an effective density of the charges that are part of a dipole. In the example of an insulating dielectric between metal capacitor plates, the only free charges are on the metal plates and dielectric contains only dipoles. The net, unbalanced bound charge at the metal/dielectric interface balances the charge on the metal plate. If the dielectric is replaced by a doped semiconductor or an ionised gas, etc, then electrons move relative to the ions, and if the system is finite they both contribute to \rho_\text{f} at the edges. D is not determined exclusively by the free charge. As E has a curl of zero in electrostatic situations, it follows that \nabla \times \mathbf{D} = \nabla \times \mathbf{P} The effect of this equation can be seen in the case of an object with a "frozen in" polarization like a bar electret, the electric analogue to a bar magnet. There is no free charge in such a material, but the inherent polarization gives rise to an electric field, demonstrating that the D field is not determined entirely by the free charge. The electric field is determined by using the above relation along with other boundary conditions on the polarization density to yield the bound charges, which will, in turn, yield the electric field. == History ==

The earliest known use of the term is from the year 1864, in James Clerk Maxwell's paper A Dynamical Theory of the Electromagnetic Field. Maxwell introduced the term D, specific capacity of electric induction, in a form different from the modern and familiar notations. It was Oliver Heaviside who reformulated the complicated Maxwell's equations to the modern form. It wasn't until 1884 that Heaviside, concurrently with Willard Gibbs and Heinrich Hertz, grouped the equations together into a distinct set. This group of four equations was known variously as the Hertz–Heaviside equations and the Maxwell–Hertz equations, and is sometimes still known as the Maxwell–Heaviside equations; hence, it was probably Heaviside who lent D the present significance it now has. == Example: Displacement field in a capacitor ==

Consider an infinite parallel plate capacitor where the space between the plates is empty or contains a neutral, insulating medium. In both cases, the free charges are only on the metal capacitor plates. Since the flux lines D end on free charges, and there are the same number of uniformly distributed charges of opposite sign on both plates, then the flux lines must all simply traverse the capacitor from one side to the other. In SI units, the charge density on the plates is proportional to the value of the D field between the plates. This follows directly from Gauss's law, by integrating over a small rectangular box straddling one plate of the capacitor: : {{Oiint|intsubscpt=\scriptstyle _A|integrand=\mathbf{D} \cdot \mathrm{d}\mathbf{A}=Q_\text{free}}} On the sides of the box, dA is perpendicular to the field, so the integral over this section is zero, as is the integral on the face that is outside the capacitor where D is zero. The only surface that contributes to the integral is therefore the surface of the box inside the capacitor, and hence |\mathbf{D}| A = |Q_\text{free}|, where A is the surface area of the top face of the box and Q_\text{free}/A=\rho_\text{f} is the free surface charge density on the positive plate. If the space between the capacitor plates is filled with a linear homogeneous isotropic dielectric with permittivity \varepsilon =\varepsilon_0\varepsilon_r, then there is a polarization induced in the medium, \mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}=\varepsilon\mathbf{E} and so the voltage difference between the plates is V =|\mathbf{E}| d =\frac{|\mathbf{D}|d}{\varepsilon}= \frac{|Q_\text{free}|d}{\varepsilon A} where d is their separation. Introducing the dielectric increases ε by a factor \varepsilon_r and either the voltage difference between the plates will be smaller by this factor, or the charge must be higher. The partial cancellation of fields in the dielectric allows a larger amount of free charge to dwell on the two plates of the capacitor per unit of potential drop than would be possible if the plates were separated by vacuum. If the distance d between the plates of a finite parallel plate capacitor is much smaller than its lateral dimensions we can approximate it using the infinite case and obtain its capacitance as C = \frac{Q_\text{free}}{V} \approx \frac{Q_\text{free}}{|\mathbf{E}| d} = \frac{A}{d} \varepsilon, == See also ==