Editing Displacement current (section)

====Current in capacitors====
An example illustrating the need for the displacement current arises in connection with [[capacitor]]s with no medium between the plates. Consider the charging capacitor in the figure. The capacitor is in a circuit that causes equal and opposite charges to appear on the left plate and the right plate, charging the capacitor and increasing the electric field between its plates. No actual charge is transported through the vacuum between its plates. Nonetheless, a magnetic field exists between the plates as though a current were present there as well. One explanation is that a ''displacement current'' {{math|''I''<sub>D</sub>}} "flows" in the vacuum, and this current produces the magnetic field in the region between the plates according to [[Ampère's law]]:<ref name=Palmer>
{{cite book
 |first1=Stuart B. |last1=Palmer
 |first2=Mircea S. |last2=Rogalski
 |name-list-style=amp
 |year=1996
 |title=Advanced University Physics  |page=214
 |publisher=Taylor & Francis
 |isbn=978-2-88449-065-8
 |url=https://books.google.com/books?id=TF6Igz5lJLgC&pg=PP1
 |via=Google Books
}}
</ref><ref name=Serway>
{{cite book
 |first1=Raymond A. |last1=Serway
 |first2=John W.    |last2=Jewett
 |name-list-style=amp
 |title=Principles of Physics
 |year=2006
 |publisher=Thomson Brooks/Cole   |page=807
 |isbn=978-0-534-49143-7
 |url=https://books.google.com/books?id=1DZz341Pp50C&pg=PA807
 |via=Google Books
}}
</ref>

[[File:Current continuity in capacitor.svg|thumb|200px | An electrically charging capacitor with an imaginary cylindrical surface surrounding the left-hand plate. Right-hand surface {{mvar|R}} lies in the space between the plates and left-hand surface {{mvar|L}} lies to the left of the left plate. No conduction current enters cylinder surface {{mvar|R}}, while current {{mvar|I}} leaves through surface {{mvar|L}}. Consistency of Ampère's law requires a displacement current {{math|1= ''I''<sub>D</sub> = ''I''}} to flow across surface {{mvar|R}}.]]

<math display=block>\oint_C \mathbf{B} \cdot \operatorname{d}\!\boldsymbol{\ell} = \mu_0 I_\mathrm{D} ~ ,</math>

where
* <math>\oint_C </math> is the closed [[line integral]] around some closed curve {{mvar|C}};
* <math>\mathbf{B} </math> is the [[magnetic field]] measured in [[tesla (unit)|teslas]];
* <math>\operatorname{\cdot} ~ </math> is the vector [[dot product]];
* <math>\mathrm{d} \boldsymbol{\ell} </math> is an [[infinitesimal]] vector line element along the curve {{mvar|C}}, that is, a vector with magnitude equal to the length element of {{mvar|C}}, and direction given by the tangent to the curve {{mvar|C}};
* <math>\mu_0 \, </math> is the [[magnetic constant]], also called the permeability of free space; and
* <math>I_\mathrm{D} \, </math> is the net displacement current that passes through a small surface bounded by the curve {{mvar|C}}.

The magnetic field between the plates is the same as that outside the plates, so the displacement current must be the same as the conduction current in the wires, that is,

<math display=block>I_\mathrm{D} = I \, ,</math>

which extends the notion of current beyond a mere transport of charge.

Next, this displacement current is related to the charging of the capacitor. Consider the current in the imaginary cylindrical surface shown surrounding the left plate. A current, say {{mvar|I}}, passes outward through the left surface {{mvar|L}} of the cylinder, but no conduction current (no transport of real charges) crosses the right surface {{mvar|R}}. Notice that the electric field {{math|'''E'''}} between the plates increases as the capacitor charges. That is, in a manner described by [[Gauss's law]], assuming no dielectric between the plates:

<math display=block>Q(t) = \varepsilon_0  \oint_S \mathbf{E}(t) \cdot \operatorname{d}\!\mathbf{S}\, ,</math>

where {{mvar|S}} refers to the imaginary cylindrical surface. Assuming a parallel plate capacitor with uniform electric field, and neglecting fringing effects around the edges of the plates, according to [[Maxwell's equations|charge conservation equation]]

<math display=block>I = -\frac{\mathrm{d} Q}{\mathrm{d} t} = - \varepsilon_0  \oint_S\frac{\partial \mathbf{E}}{\partial t} \cdot \operatorname{d}\!\mathbf{S} = S \, \varepsilon_0 \Biggl. \frac{\partial \mathbf{E}}{\partial t} \Biggr|_R  ~ , </math>

where the first term has a negative sign because charge leaves surface  {{mvar|L}} (the charge is decreasing), the last term has a positive sign because unit vector of surface {{mvar|R}} is from left to right while the direction of electric field is from right to left, {{mvar|S}} is the area of the surface {{mvar|R}}. The electric field at surface {{mvar|L}} is zero because surface {{mvar|L}} is in the outside of the capacitor. Under the assumption of a uniform electric field distribution inside the capacitor, the displacement current density '''{{math|J}}'''<sub>D</sub> is found by dividing by the area of the surface:

<math display=block> \mathbf{J}_\mathrm{D} = \frac{\mathbf{I}_\mathrm{D}}{S} = \frac{\mathbf I}{S} =  \varepsilon_0 \frac{\partial \mathbf E}{\partial t} = \frac{\partial  \mathbf D}{\partial t} ~ , </math>

where  '''{{math|I}}'''  is the current leaving the cylindrical surface (which must equal '''{{math|I}}'''<sub>D</sub>) and '''{{math|J}}'''<sub>D</sub>  is the flow of charge per unit area into the cylindrical surface through the face {{mvar|R}}.

Combining these results, the magnetic field is found using the integral form of [[Ampère's law]] with an arbitrary choice of contour provided the displacement current density term is added to the conduction current density (the Ampère-Maxwell equation):<ref name="Feynman">
{{cite book
 | first1 = Richard P. | last1 = Feynman  
 | first2 = Robert     | last2 = Leighton
 | first3 = Matthew    | last3 = Sands
 | name-list-style = amp
 | year   = 1963
 | title  = The Feynman Lectures on Physics
 | volume = 2   | page = 18‑4
 | publisher = Addison-Wesley
 | location = Massachusetts, USA
 | isbn =978-0-201-02116-5
 | url = https://archive.org/details/feynmanlectureso0003feyn
 | url-access = registration
 | via = archive.org
}}
</ref>

<math display=block>\oint_{\partial S} \mathbf{B} \cdot \operatorname{d}\!\boldsymbol{\ell} = \mu_0 \int_S \left(\mathbf{J} + \epsilon_0 \frac {\partial \mathbf{E}}{\partial t}\right) \cdot \operatorname{d}\! \mathbf{S}\,.</math>

This equation says that the integral of the magnetic field {{math|'''B'''}} around the edge {{tmath|\partial S}} of a surface {{mvar|S}} is equal to the integrated current {{math|'''J'''}} through any surface with the same edge, plus the displacement current term {{tmath|\varepsilon_0 \partial \mathbf{E} / \partial t}} through whichever surface. 
{{Clear}}
[[File:Displacement current in capacitor.svg|thumb|200px|Example showing two surfaces {{math|''S''<sub>1</sub>}} and {{math|''S''<sub>2</sub>}} that share the same bounding contour {{math|∂''S''}}. However, {{math|''S''<sub>1</sub>}} is pierced by conduction current, while {{math|''S''<sub>2</sub>}} is pierced by displacement current. Surface {{math|''S''<sub>2</sub>}} is closed under the capacitor plate.]]

As depicted in the figure to the right, the current crossing surface {{math|''S''<sub>1</sub>}} is entirely conduction current. Applying the Ampère-Maxwell equation to surface {{math|''S''<sub>1</sub>}} yields:

<math display=block>B = \frac {\mu_0 I}{2 \pi r} ~ .</math>

However, the current crossing surface {{math|''S''<sub>2</sub>}} is entirely displacement current. Applying this law to surface {{math|''S''<sub>2</sub>}}, which is bounded by exactly the same curve {{tmath|\partial S}}, but lies between the plates, produces:

<math display=block>B = \frac {\mu_0 I_\mathrm{D}}{2 \pi r} ~ .</math>

Any surface  {{math|''S''<sub>1</sub>}} that intersects the wire has current {{mvar|I}} passing through it so [[Ampère's law]] gives the correct magnetic field. However a second surface {{math|''S''<sub>2</sub>}} bounded by the same edge {{tmath|\partial S}}  could be drawn passing between the capacitor plates, therefore having no current passing through it. Without the displacement current term Ampere's law would give zero magnetic field for this surface. Therefore, without the displacement current term Ampere's law gives inconsistent results, the magnetic field would depend on the surface chosen for integration. Thus the displacement current term {{tmath|\varepsilon_0 \partial \mathbf{E} / \partial t}} is necessary as a second source term which gives the correct magnetic field when the surface of integration passes between the capacitor plates. Because the current is increasing the charge on the capacitor's plates, the electric field between the plates is increasing, and the rate of change of electric field gives the correct value for the field {{math|'''B'''}} found above.