Question

A parallel plate capacitor consists of two circular plates separated by distance and each of radius as $0.1mm,2cm$respectively. If voltage across the plates is at rate of $5\times {{10}^{13}}V/s,$displacement current value is
$\begin{align}
  & a)5.50A \\
 & b)5.56\times {{10}^{3}}A \\
 & c)2.28\times {{10}^{4}}A \\
 & d)5.56\times {{10}^{2}}A \\
\end{align}$

Answer 1

Hint: Convert all the given values into the same units. The displacement current is found out by product of area of the plates, permeability and the rate of change of voltage. Value of permittivity of free space is given by $8.85\times {{10}^{-12}}F{{m}^{-1}}$.

Formula used:
$\begin{align}
  & A=\pi {{r}^{2}} \\
 & {{I}_{d}}={{\varepsilon }_{0}}A(\dfrac{dE}{dt}) \\
\end{align}$

Complete answer:
Given the radius of the circular plates as $2cm=2\times {{10}^{-2}}m$.

Area of the plates is calculated as,
$\begin{align}
  & A=\pi {{(2\times {{10}^{-2}})}^{2}} \\
 & A=4\pi \times {{10}^{-4}} \\
\end{align}$
Now, the displacement current is calculated as,
$\begin{align}
  & {{I}_{d}}=A{{\varepsilon }_{0}}(\dfrac{dE}{dt}) \\
 & {{I}_{d}}=4\pi \times {{10}^{-4}}\times 8.85\times {{10}^{-12}}\times 5\times {{10}^{13}} \\
 & {{I}_{d}}=555.78\times {{10}^{-3}}A \\
\end{align}$

So, the correct answer is “Option A”.

Additional Information:
When a capacitor is charging or discharging, current flows in the circuit. However, there will not be any actual transfer in the insulated region between capacitors which is opposite to flow of current. Therefore, displacement current is the current in the insulated region due to the changing electric flux. The displacement current is a quantity appearing in Maxwell's equations that is defined in terms of rate of change of electric displacement field. Displacement current has the units of the electric current, and it has an associated magnetic field similar to the actual currents do. The displacement current is not a real current but has the same properties as a real current do. When there’s a changing electric field, we can treat its effects as due to displacement current density arising from field variation.

Note:
Displacement current is not real current but acts as a real current when the electric field changes. It was first added by Maxwell to the amperes current law. The direction of displacement current is the same as the direction of electric field when rate of change of field is positive and vice versa.