Equation of Continuity and Displacement Current
Equation of continuity states that “ If the net charge crossing a surface enclosing closed volume is not zero, then the charge density within the volume must charge with time in a manner that the rate of increase of charge within the volume equals the net rate of charge into the volume. ” This note provides us an information on equation of continuity.
Summary
Equation of continuity states that “ If the net charge crossing a surface enclosing closed volume is not zero, then the charge density within the volume must charge with time in a manner that the rate of increase of charge within the volume equals the net rate of charge into the volume. ” This note provides us an information on equation of continuity.
Things to Remember
Equation of continuity states that “ If the net charge crossing a surface enclosing closed volume is not zero, then the charge density within the volume must charge with time in a manner that the rate of increase of charge within the volume equals the net rate of charge into the volume. ”
Since \(\vec J’\) arise due to variation of electric displacement \(\vec D\) with time it is termed as displacement current.
According to Maxwell, it is just as effective as J, the conduction current density, in producing magnetic field.
\(\vec J’\) results in unification of electric and magnetic phenomena.
In a good conductor, \(\vec J’\) is negligible compared to \(\vec j\) at frequency lower than light frequencies \((10^{15})\).
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Equation of Continuity and Displacement Current
Equation of continuity states that “ If the net charge crossing a surface enclosing closed volume is not zero, then the charge density within the volume must charge with time in a manner that the rate of increase of charge within the volume equals the net rate of charge into the volume. ”
Let us assume that charge \(‘\rho ‘\) is a function of time. The transport of charge is responsible for current.
$$ \text {i.e.}\: I = \frac {dq}{dt} $$
If \(\rho \) is volume charge density, which is given by
\begin{align*} \rho &= \frac {dq}{dV}, \\ q &= \int \rho \: dV\\ I &= \frac {d}{dt}\int \rho \: dV\dots (1)\\\end{align*}
Here, it has been considered that the current is as the extended in space of volume V closed by a surface S. The current density J is defined as net amount of charge crossing the unit area normal to the direction of charge flow of the surface in unit time.
$$I = \int \vec J. d\vec s, \dots (2) $$
From equations (1) and (2), it becomes,
$$ \int \vec J. d\vec s = \frac {d}{dt} \int \rho \: dV$$
The current is flowing outward direction so current I should be negative and hence above relation becomes,
\begin{align*} \int \vec J. d\vec s &= -\frac {d}{dt}\int \rho \: dV \\ \text {Gauss divergence theorem,}\\ \int \vec J. d\vec s &= \int (\nabla .\vec J)d\vec V \\ \int (\nabla .\vec J)d\vec V &= - \frac {d}{dt}\rho \: dV \\ \text {or,}\: \int \left (\nabla .\vec J + \frac {d\rho }{dt} \right )d\vec V &= 0 \end{align*}
Since, the arbitrary volume \(dV\neq 0,\)
$$\nabla .\vec J + \frac {d\rho }{dt}= 0\dots ( 3) $$
This equation (3) is called the equation of continuity. For stationary current \(\rho \) is constant. So, \(\frac {d\rho}{dt} =0\). Hence,
$$\nabla . \vec J = 0$$
DisplacementCurrent
We know the ampere’s circuital law which states that, the integral of the magnetic field intensity in a closed loop is equal to the \(\mu_0\) time the current in that loop.
\begin{align*} \text {i.e.}\: \oint \vec B.d\vec l &= \mu_oI \\ \text {or,}\: \oint \vec H.d\vec l &= I \dots (1) \\ \text {We know,}\: I &= \int_s \vec J. d\vec s \dots (2) \\ \therefore \int \vec H. d\vec l &= \int_s \vec J. d\vec s \\ \text {From stroke}’s\: \text {theorem,} \\ \oint \vec H.d\vec l &= \int_s(\nabla \times \vec H).d\vec s \\ \text {so,}\: \int (\nabla \times \vec H). ds &= \int_s \vec J.d\vec s \\ \text {or,}\: \int (\nabla \times H – J).d\vec s = 0 \:\:\:[\text {since}\:d\vec s \neq 0]\\ \nabla \times \vec H &= \vec J \times (3) \\ \text {Taking divergence, then} \\ \nabla .(\nabla \times \vec H) &= \vec {\nabla }. \vec J\\ 0 &= -\frac {\partial \rho}{\partial t} \end{align*}
Which shows that, the equation (3) is true for steady state conditions in which charge density is not changing. Equation (3) must be modified for time dependent fields. Maxwell thought that definition of current density is incomplete and added something to J. So that equation (3) becomes
\begin{align*} \text {curl}\: \vec H &= J + J’ \\ \text {On taking divergene,}\\ \nabla .(\nabla \times H) &= \nabla .(\vec J + \vec J’) \\ 0 &= \nabla .\vec J + \nabla . \vec {J’} \\ \text {or,}\: \text {div}\: \vec {J’} &= - \text {div} \: \vec J\:\:\:[\because \nabla .\vec {J’} = \nabla . \ vec J] \\ &= \frac {\partial \rho}{\partial t} \end{align*}
We know from gauss theorem \(\nabla . \vec D = \rho \). So from equation (4)
\begin{align*} \nabla . \vec {J}' &= \frac {\partial }{\partial t}(\nabla . \vec D) = \nabla .\frac {\partial \vec D}{\partial t} \\ \vec {J}' &= \frac {\partial \vec D}{\partial t} \\ \text {hence,}\: \nabla \times \vec H= \vec J + \vec {J}' &= J + \frac {\partial \vec D}{\partial t} \dots (5) \end{align*}
Note:
- Since \(\vec J’\) arise due to variation of electric displacement \(\vec D\) with time it is termed as displacement current. According to Maxwell, it is just as effective as J, the conduction current density, in producing magnetic field.
- \(\vec J’\) results in unification of electric and magnetic phenomena.
- In a good conductor, \(\vec J’\) is negligible compared to \(\vec j\) at frequency lower than light frequencies \((10^{15})\).
Bibliography
P.B. Adhikari, Bhoj Raj Gautam, Lekha Nath Adhikari. A Textbook of Physics. kathmandu: Sukunda Pustak Bhawan, 2011.
Jha, V. K.; 'Lecture title'; Maxwell's Equation; St. Xavier's College, Kathmandu; 2016.
Lesson
Maxwell's Electromagnetic Equations
Subject
Physics
Grade
Bachelor of Science
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