Question

Show that the dimensions of the displacement current ${\varepsilon _0}\dfrac{{d{\phi _\varepsilon }}}{{dt}}$ are that of an electric current.

Answer 1

Hint: In this solution, we will first find the dimensional formula of permittivity and then of the electric flux. Then we will use their dimensional formula to determine the dimensions of displacement current.

Complete step by step answer:
Let us start by finding the dimensional formula of permittivity. We know that the force between two charged particles can be written as
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Then we can write the permittivity as
Now the dimensional formula that we will need in the above equation are $[F] = {M^1}{L^1}{T^{ - 2}}$, $[{q_1}] = [{q_2}] = [{A^1}{T^1}]$, $[r] = {M^0}{L^1}{T^0}$
So, we can get the dimensional formula of permittivity as
$[{\varepsilon _0}] = {M^{ - 1}}{L^{ - 3}}{T^4}{A^2}$
Now let us find the dimensional formula of electric flux using Gauss’s law, we can write that
$\phi = E.A$
Since the dimensions of the electric field can be determined from$F = qE$ as
$[E] = {M^1}{L^1}{T^{ - 4}}{A^{ - 1}}$
So, the dimensions of electric flux will be
\[[\phi ] = {M^1}{L^1}{T^{ - 3}}{A^{ - 1}} \times {L^2}\]
\[ \Rightarrow [\phi ] = {M^1}{L^3}{T^{ - 3}}{A^{ - 1}}\]
Then the dimensions of displacement current will be equal to
$\left[ {{\varepsilon _0}\dfrac{{d{\phi _\varepsilon }}}{{dt}}} \right] = {M^{ - 1}}{L^{ - 3}}{T^4}{A^2} \times {M^1}{L^3}{T^{ - 4}}{A^{ - 1}}$
$\left[ {{\varepsilon _0}\dfrac{{d{\phi _\varepsilon }}}{{dt}}} \right] = {A^1}$
Which are the dimensions of electric current. Hence we have proved that the dimension of displacement current is equal to the dimensional formula of current.

Note: Displacement current is an important aspect of Maxwell’s equations which was added later as a correction to the four Maxwell’s laws of electromagnetism. The need for it arises because of the following reasoning. When we charge a capacitor, there is a current flowing between the plates, but there is no transfer of charges in the insulating region between the capacitor plates. This contradicts the understanding of electric current and can be explained by the concept of displacement current. Therefore, we can say that there is an electric current in the region due to the changing electric flux between the capacitors.