Question

How rapidly should the electric flux through an area change to create a displacement current of $0.50\,mA$ ?

Answer 1

Hint: In electrostatics, electric flux is the amount of electric field lines passing through a given area and displacement current is the current which passes between the plates of a charging capacitor.

 Formula used:
Displacement current is calculated by using the formula,
${I_d} = { \in _0}\dfrac{{d{\phi _E}}}{{dt}}$
where ${ \in _0} = 8.854 \times {10^{ - 12}}{C^2}{m^{ - 2}}{N^{ - 1}}$ which is known as permittivity of free space and $d{\phi _E}$ is the amount of electric flux change in the time $dt.$

Complete step by step answer:
Let us first understand basic of ampere circuital law, which states that line integral of magnetic field is equal to ${\mu _0}$ 9permiability of free space) times the current across the conductor but later Maxwell introduced that a displacement current also flows between the plates of capacitor and has a magnitude of ${I_d} = { \in _0}\dfrac{{d{\phi _E}}}{{dt}}$ and then the complete ampere circuital law can be written mathematically as,
$\int {\vec B.d\vec l = {\mu _0}I + {\mu _0}{ \in _0}\dfrac{{d{\phi _E}}}{{dt}}} $

Now, according to the question in order to create a displacement current of ${I_d} = 0.50mA = 0.5 \times {10^{ - 3}}A$ the rate of change of electric flux is calculated but using the formula ${I_d} = { \in _0}\dfrac{{d{\phi _E}}}{{dt}}$ putting the values we get,
$\dfrac{{{I_d}}}{{{ \in _0}}} = \dfrac{{d{\phi _E}}}{{dt}}$
$ \Rightarrow \dfrac{{d{\phi _E}}}{{dt}} = \dfrac{{0.5 \times {{10}^{ - 3}}}}{{8.854 \times {{10}^{ - 12}}}}$
$ \Rightarrow\dfrac{{d{\phi _E}}}{{dt}} = 0.056 \times {10^9}$
$\therefore \dfrac{{d{\phi _E}}}{{dt}} = 5.6 \times {10^7}\,V{m^{ - 1}}{s^{ - 1}}$

Hence, the rate at which electric flux need to change is $\dfrac{{d{\phi _E}}}{{dt}} = 5.6 \times {10^7}V{m^{ - 1}}{s^{ - 1}}$.

Note: It should be remembered that, Maxwell found that whenever the charge inside the closed surface is change which means electric flux is changing we need to add some other form of current too in order to validate Ampere circuital law, this led to the discovery of displacement current and later this whole equation $\int {\vec B.d\vec l = {\mu _0}I + {\mu _0}{ \in _0}\dfrac{{d{\phi _E}}}{{dt}}} $ is known as one of the four Maxwell equations of electromagnetism and basic unit of conversion is used as $1mA = {10^{ - 3}}A.$