91影视

When the current flowing in a conductor is because of the 鈥榝low of charges鈥�, then the current is defined as 鈥榗onduction current鈥�.

While the current flow in a conductor because of the fluctuations in the electric field is defined as the 鈥榙isplacement current鈥�.

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.

The sea water frequency is, $谓=4脳{10}^{8}Hz.$

The sea water permittivity is, $鈭�=81{鈭�}_0.$

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

The sea water resistivity is, $蚁=0.23惟.m.$

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

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,

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

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

$$\begin{gathered} J_c=\frac{E}{蚁} \\ {J}_c=\frac{V_0\cos 2\mathrm{蟺谓迟}}{蚁d} \\ {J}_c=\left(\frac{V_0}{蚁d}\right)\cos 2\mathrm{蟺谓迟} \\ {J}_c={\left(j_0\right)}_c\cos 2\mathrm{蟺谓迟} \end{gathered}$$

Here, role="math" localid="1657533906249" $\left(\frac{V_0}{蚁d}\right)={\left(J_0\right)}_c$, and${\left(J_0\right)}_c$ is the amplitude of the conduction current density.

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

$$\begin{gathered} J_d=鈭�\frac{d}{dt}\left(\frac{V_0\cos 2\mathrm{蟺谓迟}}{d}\right) \\ {J}_d=\frac{-2\mathrm{蟺谓}鈭�}{d}\left(V_0\sin 2\mathrm{蟺谓迟}\right) \\ {\mathrm{J}}_{\mathrm{d}}=\left(\frac{-2\mathrm{蟺谓}鈭�{\mathrm{V}}_0}{\mathrm{d}}\right)\sin 2\mathrm{蟺谓迟} \\ {\mathrm{J}}_{\mathrm{d}}={\left({\mathrm{J}}_0\right)}_{\mathrm{d}}\sin 2\mathrm{蟺谓迟} \end{gathered}$$

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

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

$$\begin{gathered} \frac{{\left(J_0\right)}_d}{{\left(J_0\right)}_c}=\frac{\left(\frac{-2\mathrm{蟺谓}鈭�{\mathrm{V}}_0}{D}\right)}{\left({\displaystyle \frac{V_0}{蚁d}}\right)} \\ \frac{{\left(J_0\right)}_d}{{\left(J_0\right)}_c}=\frac{-2\mathrm{蟺谓}鈭�{\mathrm{V}}_0}{D}脳\frac{蚁d}{V_0} \\ \frac{{\left(J_0\right)}_d}{{\left(J_0\right)}_c}=-2\mathrm{蟺谓}鈭�蚁 \\ \frac{{\left(J_0\right)}_d}{{\left(J_0\right)}_c}=2\mathrm{蟺}(4脳{10}^8\mathrm{Hz})(81{鈭�}_0)(0.23\mathrm{惟}.\mathrm{m}) \end{gathered}$$

Solve further as:

$\frac{{\left(J_0\right)}_d}{{\left(J_0\right)}_c}=-(4\mathrm{蟺}{鈭�}_0)脳(2脳{10}^8脳81脳0.23)\mathrm{Hz}.\mathrm{惟}.\mathrm{m}$

Substitute the value$(4\mathrm{蟺}{鈭�}_0)=\frac{1}{(9脳{10}^9){\mathrm{Nm}}^2/{\mathrm{C}}^2}.$ ,

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

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.