Question

What is displacement current? Obtain an expression of displacement current for a charged capacitor. Write Ampere- Maxwell’s law.

Answer 1

We can derive an expression for the displacement current for a charged capacitor by using the formula for electric field and then differentiating it, we can get the displacement current. For the Ampere-Maxwell’s law, we can use the actual Ampere law and can modify it to get the Ampere-Maxwell’s law.
Formula used:

& E=\dfrac{\sigma }{{{\varepsilon }_{0}}} \\\ & \sigma =\dfrac{Q}{A} \\\ & \nabla \times B={{\mu }_{0}}J \\\ & \nabla \times B={{\mu }_{0}}(J+{{J}_{D}}) \\\ \end{aligned}$$ **Complete answer:** Displacement is the current generated due to the changing electric displacement field and it describes how magnetic fields are produced or acts when the electric field is changing.  Let us consider a charging capacitor, then the electric field is given as $$E=\dfrac{\sigma }{{{\varepsilon }_{0}}}$$ ………………..(i) Where σ is the surface charge density and $${{\varepsilon }_{0}}$$ is the permittivity in the free space. Surface charge density of the capacitor given as $$\sigma =\dfrac{Q}{A}$$ Where Q is the charge on the plates and A is its area. Substituting the value of surface charge density in equation (i), $$E=\dfrac{Q}{{{\varepsilon }_{0}}A}$$ To get the electric field between the plates can be obtained by differentiating the electric field with respect to time. $$\dfrac{\partial E}{\partial t}=\dfrac{1}{{{\varepsilon }_{0}}A}\dfrac{dQ}{dt}=\dfrac{1}{{{\varepsilon }_{0}}A}{{I}_{D}}$$ $$\begin{aligned} & \implies{{I}_{D}}={{\varepsilon }_{0}}A\dfrac{\partial E}{\partial t} \\\ & \implies{{I}_{D}}={{\varepsilon }_{0}}\dfrac{d\phi }{dt} \\\ \end{aligned}$$ Where $${{I}_{D}}$$ is the displacement current. Hence the current between the plates of the capacitor is not zero instead there is this displacement current flowing through it. Displacement current is the cause of energy transport in the capacitor. The Ampere law is given as $$\nabla \times B={{\mu }_{0}}J$$ Where, B is the magnetic field, J is current density and $${{\mu }_{0}}$$is permeability in a free space. But according to Maxwell the Ampere law was not valid if the field was dynamic. Hence he added a new term in Ampere law such that electromagnetic fields are also symmetric. The modified equation is $$\begin{aligned} & \nabla \times B={{\mu }_{0}}J+{{\mu }_{0}}{{\varepsilon }_{0}}\dfrac{\partial E}{\partial t} \\\ & \implies\nabla \times B={{\mu }_{0}}J+{{\mu }_{0}}{{J}_{D}} \\\ & \implies\nabla \times B={{\mu }_{0}}(J+{{J}_{D}}) \\\ \end{aligned}$$ **Where, $${{J}_{D}}$$ is the current displacement and it is given as $${{J}_{D}}={{\varepsilon }_{0}}\dfrac{\partial E}{\partial t}$$** This modified Ampere’s law is known as Ampere-Maxwell’s law. **Note:** In the capacitor, the current flowing outside the plate is the conductive current and the current between the plates is the displacement current. Displacement current arises due to the accumulation of charge on the plate. Here, we used two notations for displacement current, in case of capacitor we used $${{I}_{D}}$$ and in case of Ampere-Maxwell’s law we used $${{J}_{D}}$$, but both are displacement current only the different notations are used just to keep the equation same as it is used in textbooks.