Q40P - Introduction to Electrodynamics

Question

Sea water at frequency $\mathbf{v}\mathbf{=}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{Hz}$ has permittivity $\mathbf{\in }\mathbf{=}\mathbf{81}\mathbf{ }{\mathbf{\in }}_{\mathbf{0}}$ , permeability $\mathbf{\mu }\mathbf{=}{\mathbf{\mu }}_{\mathbf{0}}$ , and resistivity $\mathbf{\rho }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{23}\mathbf{\Omega }\mathbf{.}\mathbf{m}$ . What is the ratio of conduction current to displacement current? [Hint: Consider a parallel-plate capacitor immersed in sea water and driven by a voltage ${\mathbf{V}}_{\mathbf{0}}\mathbf{cos}\mathbf{\left(}\mathbf{2}\mathbf{\pi vt}\mathbf{\right)}$ .]

Step-by-Step Solution

Answer

The value of the ratio is 2.41.

Step 1: Conduction current and Displacement current

When the current flowing in a conductor is because of the ‘flow of charges’, then the current is defined as ‘conduction current’.

While the current flow in a conductor because of the fluctuations in the electric field is defined as the ‘displacement current’.

 

Based on the definition, the conduction current can be determined by using Ohm's Law while the displacement current doesn't follow Ohm's Law.

Step 2: Given information

The sea water frequency is, $\nu =4\times {10}^8 Hz.$

The sea water permittivity is, $\in =81 {\in }_0.$

The sea water permeability is, $\mu ={\mu }_0.$

The sea water resistivity is, $\rho =0.23\Omega .m.$

The voltage of the parallel-plate capacitor immersed in sea water is, $V\left(t\right)=V_0\cos \left(2\mathrm{\pi \nu t}\right).$

Step 3: The conduction current

Assume, the distance between the two plates of a parallel plate capacitor is d.

 

The formula for the electric field between two plates of a parallel plate capacitor is given by,

 

$$\begin{gathered} E=\frac{V\left(t\right)}{d} \\ E=\frac{V_0\cos 2\mathrm{\pi \nu t}}{d} \end{gathered}$$

 

Then, the formula for the conduction current density of the capacitor is given by,

 

$$\begin{gathered} J_c=\frac{E}{\rho } \\ {J}_c=\frac{V_0\cos 2\mathrm{\pi \nu t}}{\rho d} \\ {J}_c=\left(\frac{V_0}{\rho d}\right)\cos 2\mathrm{\pi \nu t} \\ {J}_c={\left(j_0\right)}_c\cos 2\mathrm{\pi \nu t} \end{gathered}$$

 

Here, $\left(\frac{V_0}{\rho d}\right)={\left(J_0\right)}_c$, and ${\left(J_0\right)}_c$ is the amplitude of the conduction current density.

Step 4: The displacement current

The formula for the displacement current density of the capacitor is given by,

 

$$\begin{gathered} J_d=\in \frac{d}{dt}\left(\frac{V_0\cos 2\mathrm{\pi \nu t}}{d}\right) \\ {J}_d=\frac{-2\mathrm{\pi \nu }\in }{d}\left(V_0\sin 2\mathrm{\pi \nu t}\right) \\ {\mathrm{J}}_{\mathrm{d}}=\left(\frac{-2\mathrm{\pi \nu }\in {\mathrm{V}}_0}{\mathrm{d}}\right)\sin 2\mathrm{\pi \nu t} \\ {\mathrm{J}}_{\mathrm{d}}={\left({\mathrm{J}}_0\right)}_{\mathrm{d}}\sin 2\mathrm{\pi \nu t} \end{gathered}$$

Here, $\frac{-2\mathrm{\pi \nu } \in {\mathrm{V}}_0}{d}={\left(J_0\right)}_d$, and ${\left(J_0\right)}_d$ is the amplitude of the displacement current density.

Step 5: The ratio of conduction current to displacement current

The ratio of conduction current to displacement current is given by,

 

$$\begin{gathered} \frac{{\left(J_0\right)}_d}{{\left(J_0\right)}_c}=\frac{\left(\frac{-2\mathrm{\pi \nu }\in {\mathrm{V}}_0}{D}\right)}{\left({\displaystyle \frac{V_0}{\rho d}}\right)} \\ \frac{{\left(J_0\right)}_d}{{\left(J_0\right)}_c}=\frac{-2\mathrm{\pi \nu }\in {\mathrm{V}}_0}{D}\times \frac{\rho d}{V_0} \\ \frac{{\left(J_0\right)}_d}{{\left(J_0\right)}_c}=-2\mathrm{\pi \nu }\in \rho \\ \frac{{\left(J_0\right)}_d}{{\left(J_0\right)}_c}=2\mathrm{\pi }(4\times {10}^8\mathrm{Hz})(81 {\in }_0)(0.23 \mathrm{\Omega }.\mathrm{m}) \end{gathered}$$

 

Solve further as:

$\frac{{\left(J_0\right)}_d}{{\left(J_0\right)}_c}=-(4\mathrm{\pi } {\in }_0)\times (2\times {10}^8\times 81\times 0.23)\mathrm{Hz}.\mathrm{\Omega }.\mathrm{m}$

 

Substitute the value $(4\mathrm{\pi } {\in }_0)=\frac{1}{(9\times {10}^9){\mathrm{Nm}}^2/{\mathrm{C}}^2}.$ ,

 

$$\begin{gathered} \frac{{\left(J_0\right)}_d}{{\left(J_0\right)}_c}=\frac{37.26\times {10}^{8}Hz.\Omega .m}{(9\times {10}^9)N{m}^2/C^2} \\ \frac{{\left(J_0\right)}_d}{{\left(J_0\right)}_c}=\frac{37.26\times {10}^8}{9\times {10}^9} \\ \frac{{\left(J_0\right)}_d}{{\left(J_0\right)}_c}=\frac{3.726}{9} \end{gathered}$$

 

Solve further as,

$\frac{{\left(J_0\right)}_d}{{\left(J_0\right)}_c}=\frac{1}{2.41}$

 

Hence, the ratio of conduction current to displacement current is 2.41.