Answer
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Hint: Using the mathematical relation for electric field between the plates of the charged capacitor, find the charge on the plates of the capacitor. Then with the help of this relation, find displacement current. Then apply ampere's circuital law and define Ampere Maxwell’s law.
Complete step-by-step answer:
When a capacitor charges or discharges, a current flow in the circuit due to the movement of the charges. But, an actual charge transfer between the insulated region between capacitors does not take place which is not a flow of current. Therefore, the displacement current is the current in the region between the capacitor plates due to the changing electric flux.
For a charged capacitor with charge Q, area of cross-section A, electric field E and the electrical flux $\phi $ the following relations hold:
$\begin{align}
& E=\dfrac{Q}{{{\varepsilon }_{o}}A} \\
& Q={{\varepsilon }_{o}}\phi \\
\end{align}$
Displacement current, ${{i}_{d}}$, can be expressed as:
${{i}_{d}}=\dfrac{dQ}{dt}={{\varepsilon }_{o}}\dfrac{d\phi }{dt}$
As a current is the charge flow per unit time.
Therefore, the expression for the displacement current becomes:
${{i}_{d}}={{\varepsilon }_{o}}\dfrac{d\phi }{dt}$
Ampere -Maxwell's law states that the line integral of magnetic field along a closed path is proportional to the total current from wires enclosed in it. The constant of proportionality is ${{\mu }_{o}}$.
\[\oint{{\vec{B}}}\cdot d\vec{l}={{\mu }_{o}}i\]
Where, B is the magnetic field, and i is the total current which constitutes of conducting current ${{i}_{c}}$ and displacement current ${{i}_{d}}$ .
Therefore, the expression for the law looks like:
\[\oint{{\vec{B}}}\cdot d\vec{l}={{\mu }_{o}}\left( {{i}_{c}}+{{i}_{d}} \right)={{\mu }_{o}}\left( {{i}_{c}}+{{\varepsilon }_{o}}\dfrac{d\phi }{dt} \right)\]
The Ampere-Maxwell equation relates electric currents and magnetic flux.
Note: Displacement current is due to the changing electric flux. Keep in mind that in Ampere's circuital law, total current that is the sum of conduction current and displacement current is considered. Also, when a closed loop encloses a number of conducting wires, the total current enclosed is calculated by considering the current in the opposite direction as cancelling each other. That is, the direction of current is also taken into consideration.
Complete step-by-step answer:
When a capacitor charges or discharges, a current flow in the circuit due to the movement of the charges. But, an actual charge transfer between the insulated region between capacitors does not take place which is not a flow of current. Therefore, the displacement current is the current in the region between the capacitor plates due to the changing electric flux.
For a charged capacitor with charge Q, area of cross-section A, electric field E and the electrical flux $\phi $ the following relations hold:
$\begin{align}
& E=\dfrac{Q}{{{\varepsilon }_{o}}A} \\
& Q={{\varepsilon }_{o}}\phi \\
\end{align}$
Displacement current, ${{i}_{d}}$, can be expressed as:
${{i}_{d}}=\dfrac{dQ}{dt}={{\varepsilon }_{o}}\dfrac{d\phi }{dt}$
As a current is the charge flow per unit time.
Therefore, the expression for the displacement current becomes:
${{i}_{d}}={{\varepsilon }_{o}}\dfrac{d\phi }{dt}$
Ampere -Maxwell's law states that the line integral of magnetic field along a closed path is proportional to the total current from wires enclosed in it. The constant of proportionality is ${{\mu }_{o}}$.
\[\oint{{\vec{B}}}\cdot d\vec{l}={{\mu }_{o}}i\]
Where, B is the magnetic field, and i is the total current which constitutes of conducting current ${{i}_{c}}$ and displacement current ${{i}_{d}}$ .
Therefore, the expression for the law looks like:
\[\oint{{\vec{B}}}\cdot d\vec{l}={{\mu }_{o}}\left( {{i}_{c}}+{{i}_{d}} \right)={{\mu }_{o}}\left( {{i}_{c}}+{{\varepsilon }_{o}}\dfrac{d\phi }{dt} \right)\]
The Ampere-Maxwell equation relates electric currents and magnetic flux.
Note: Displacement current is due to the changing electric flux. Keep in mind that in Ampere's circuital law, total current that is the sum of conduction current and displacement current is considered. Also, when a closed loop encloses a number of conducting wires, the total current enclosed is calculated by considering the current in the opposite direction as cancelling each other. That is, the direction of current is also taken into consideration.
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