Editing Polarization density (section)

==Gauss's law for the field of ''P''==

For a given volume {{mvar|V}} enclosed by a surface {{mvar|S}}, the bound charge <math>Q_b</math> inside it is equal to the flux of {{math|'''P'''}} through {{mvar|S}} taken with the negative sign, or

{{NumBlk|:|{{oiint|preintegral = <math>-Q_b = </math> | intsubscpt = <math>{\scriptstyle S}</math> | integrand = <math>\mathbf{P} \cdot \mathrm{d}\mathbf{A}</math>}}|{{EquationRef|3}}}}

{{math proof|proof=
Let a surface area {{mvar|S}} envelope part of a dielectric. Upon polarization negative and positive bound charges will be displaced. Let {{math|''d''<sub>1</sub>}} and {{math|''d''<sub>2</sub>}} be the distances of the bound charges <math>\mathrm d q_b^-</math> and <math>\mathrm d q_b^+</math>, respectively, from the plane formed by the element of area d''A'' after the polarization. And let {{math|d''V''<sub>1</sub>}} and {{math|d''V''<sub>2</sub>}} be the volumes enclosed below and above the area d''A''.

[[File:Surface Integral Polarization.jpg|thumb|Above: an elementary volume d''V'' = d''V<sub>1</sub>''+ d''V<sub>2</sub>'' (bounded by the element of area d'''A''') so small, that the dipole enclosed by it can be thought as that produce by two elementary opposite charges. Below, a planar view (click in the image to enlarge).]]

It follows that the negative bound charge <math>\mathrm d q_b^- = \rho_b^-\ \mathrm d V_1 = \rho_b^- d_1\ \mathrm d A</math> moved from the outer part of the surface d''A'' inwards, while the positive bound charge <math>\mathrm d q_b^+ = \rho_b\ \mathrm d V_2 = \rho_b d_2\ \mathrm d A</math> moved from the inner part of the surface outwards.

By the law of conservation of charge the total bound charge <math>\mathrm d Q_b</math> left inside the volume <math>\mathrm d V</math> after polarization is:

<math display="block">\begin{align}
\mathrm{d} Q_b
& = \mathrm{d} q_\text{in} - \mathrm{d} q_\text{out} \\
& = \mathrm{d} q_b^- - \mathrm{d} q_b^+ \\
& = \rho_b^- d_1\ \mathrm{d} A - \rho_b d_2\ \mathrm{d} A
\end{align}</math>

Since <math display="block">\rho_b^- = -\rho_b</math> and (see image to the right)
<math display="block">\begin{align}
d_1 &= (d - a)\cos(\theta) \\
d_2 &= a\cos(\theta)
\end{align}</math>

The above equation becomes
<math display="block">\begin{align}
\mathrm{d} Q_b
&= - \rho_b (d - a)\cos(\theta)\ \mathrm{d} A - \rho_b a\cos(\theta)\ \mathrm{d} A \\
&= - \rho_b d\ \mathrm{d} A \cos(\theta) 
\end{align}</math>

By ({{EquationNote|2}}) it follows that <math>\rho_b d = P</math>, so we get:
<math display="block">\begin{align}
   \mathrm{d} Q_b &= - P\ \mathrm{d} A \cos(\theta) \\
  -\mathrm{d} Q_b &= \mathbf{P} \cdot \mathrm{d} \mathbf{A}
\end{align}</math>

And by integrating this equation over the entire closed surface ''S'' we find that
:{{oiint
 | preintegral = <math>-Q_b = </math>
 | intsubscpt = <math>\scriptstyle{S}</math>
 | integrand = <math>\mathbf{P} \cdot \mathrm{d}\mathbf{A}</math>
}}

which completes the proof.
}}

===Differential form===

By the divergence theorem, Gauss's law for the field '''P''' can be stated in ''differential form'' as:
<math display="block">-\rho_b = \nabla \cdot \mathbf P,</math>
where {{math|∇ · '''P'''}} is the divergence of the field '''P''' through a given surface containing the bound charge density <math>\rho_b</math>.

{{math proof|proof=
By the divergence theorem we have that
<math display="block">-Q_b = \iiint_V \nabla \cdot \mathbf P\ \mathrm{d} V,</math>
for the volume ''V'' containing the bound charge <math>Q_b</math>. And since <math>Q_b</math> is the integral of the bound charge density <math>\rho_b</math> taken over the entire volume ''V'' enclosed by ''S'', the above equation yields
<math display="block">-\iiint_V \rho_b \ \mathrm{d} V = \iiint_V \nabla \cdot \mathbf{P}\ \mathrm{d} V ,</math>
which is true if and only if
<math>-\rho_b = \nabla \cdot \mathbf{P}</math>
}}