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Question: How rapidly should the electric flux through an area change to create a displacement current of \(0....

How rapidly should the electric flux through an area change to create a displacement current of 0.50 mA0.50\,mA ?

Explanation

Solution

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,
Id=∈0dϕEdt{I_d} = { \in _0}\dfrac{{d{\phi _E}}}{{dt}}
where ∈0=8.854×10−12C2m−2N−1{ \in _0} = 8.854 \times {10^{ - 12}}{C^2}{m^{ - 2}}{N^{ - 1}} which is known as permittivity of free space and dϕEd{\phi _E} is the amount of electric flux change in the time dt.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 μ0{\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 Id=∈0dϕEdt{I_d} = { \in _0}\dfrac{{d{\phi _E}}}{{dt}} and then the complete ampere circuital law can be written mathematically as,
∫B⃗.dl⃗=μ0I+μ0∈0dϕEdt\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 Id=0.50mA=0.5×10−3A{I_d} = 0.50mA = 0.5 \times {10^{ - 3}}A the rate of change of electric flux is calculated but using the formula Id=∈0dϕEdt{I_d} = { \in _0}\dfrac{{d{\phi _E}}}{{dt}} putting the values we get,
Id∈0=dϕEdt\dfrac{{{I_d}}}{{{ \in _0}}} = \dfrac{{d{\phi _E}}}{{dt}}
⇒dϕEdt=0.5×10−38.854×10−12\Rightarrow \dfrac{{d{\phi _E}}}{{dt}} = \dfrac{{0.5 \times {{10}^{ - 3}}}}{{8.854 \times {{10}^{ - 12}}}}
⇒dϕEdt=0.056×109\Rightarrow\dfrac{{d{\phi _E}}}{{dt}} = 0.056 \times {10^9}
∴dϕEdt=5.6×107 Vm−1s−1\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 dϕEdt=5.6×107Vm−1s−1\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 ∫B⃗.dl⃗=μ0I+μ0∈0dϕEdt\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=10−3A.1mA = {10^{ - 3}}A.