Question

Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere’s circuital law to include the term due to displacement current.

Answer 1

In order to solve this question you have to recall Ampere’s circuital law. Also, keep in mind that the space between the capacitors consists of an insulator. Also, remember the formula for the electric flux across a parallel plate capacitor.

Complete step by step solution:
Ampere’s circuital law states that the closed line integral of the magnetic field surrounding a current-carrying conductor is the absolute permeability of medium times the total current moving through the conductor.
Mathematically,
$\oint {\vec B} .\vec dl = {\mu _0}i$ ……….(i)
Now, the electric flux across a parallel plate capacitor is given by,
${\phi _E} = \dfrac{q}{{{\varepsilon _0}}}$ ………(ii)
Now, we know that the current in the parallel plate of the capacitor is given by,
$i = \dfrac{{dq}}{{dt}}$
Now on differentiating equation (ii) with respect to t, we have
$\dfrac{{d{\phi _E}}}{{dt}} = \dfrac{d}{{dt}}\left( {\dfrac{q}{{{\varepsilon _0}}}} \right)$
On further solving we have
$\Rightarrow \dfrac{{d{\phi _E}}}{{dt}} = \dfrac{1}{{{\varepsilon _0}}}\dfrac{{dq}}{{dt}}$
From this we get the value of current in the parallel plate of the capacitor as
$i = \dfrac{{dq}}{{dt}} = {\varepsilon _0}\dfrac{{d{\phi _E}}}{{dt}}$
This term is the missing term in the equation (i) of the ampere’s circuital law and this is also the displacement current in the conductor.
Hence, the total current in the conductor is the sum of displacement current and the conduction current, that is
$i = {i_c} + {i_d}$
On putting the value, we get
$\Rightarrow i = {i_c} + {\varepsilon _0}\dfrac{{d{\phi _E}}}{{dt}}$
Now, putting the that value in the equation (i) that is ampere’s circuital law, we have
$\oint {\vec B} .\vec dl = {\mu _0}{i_c} + {\mu _0}{\varepsilon _0}\dfrac{{d{\phi _E}}}{{dt}}$
This is the required generalized form of Ampere’s circuital law.

Note: Ampere’s law states that the magnetic field due to an electric current is directly proportional to the size of the electric current with a constant of proportionality equal to the permeability of free space. Always remember that the magnetic field lowers in magnitude as we move wider. Hence, this law can also be applied to calculate the magnetic field around the wire.