Correct Answer - Option 1 : 45 MHz
Concept:
Conduction current density:
- Conduction current comes from the movement of charges such as electrons present in conductors or and semiconductors.
- The electrons present in the conduction band of conductors or semiconductors are free and they respond to the applied electric field.
- The current passing through due to such free election per unit cross-sectional area per unit time is commonly referred to as conduction current density.
\({J_c} = \frac{{{I_c}}}{A} = \sigma E\left( {\frac{{Amp}}{{{m^2}}}} \right)\)
Displacement current density:
- Displacement current comes from the change in the electric field caused by the displacement of the center of positive charge and center of negative charge of a dielectric or insulating medium.
- Dielectric does not contain free electrons for conduction, rather they contain immobile charges and are displaced slightly from their equilibrium position in the presence of the electric field.
- Consequently, they produce a change in the electric field present in the dielectric medium and hence displacement current.
- The rate of change of such electric field with respect to time is commonly referred to as displacement current density.
\({J_D} = \frac{{{J_D}}}{A} = \frac{{\frac{{\partial \mu }}{{\partial t}}}}{A} = \frac{\partial }{{\partial t}}\left( {\frac{\psi }{A}} \right) = \frac{{\partial D}}{{\partial t}}\)
\({J_D} = \frac{{\partial D}}{{\partial t}} = \omega D = 2\pi f{\varepsilon _o}{\varepsilon _r}E\)
Calculation:
Given, conductivity (σ) = 10-2 mho/m
Relative permittivity (εr) = 4
Jc = σE
\({J_D} = \frac{{\partial D}}{{\partial t}} = \omega D = \omega {\varepsilon _o}{\varepsilon _r}E\)
Both currents are equal i.e. JC = JD
σ E = ωεoεrE
\(\omega = \frac{\sigma }{{{\varepsilon _o}{\varepsilon _r}}}\)
\(f = \frac{{{{10}^{ - 2}}}}{{2\pi \times 8.85 \times {{10}^{ - 12}} \times 4}} = 45\;MHz\)
