Sea water at frequency `v=4 xx 10^(8)` Hz has permittivity `epsilon ~~ 80 epsilon_(0)` permeability `mu= mu_(0)` and resistivity `rho= 0.25`M. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source `V(t) =V_(0) " sin "(2pivt).` What fraction of the conduction current density is the displacement current density ?

Options

A. `2//3`
B. `4//9`
C. `9//4`
D. `2`

Explanation

To solve the problem, we need to find the fraction of the conduction current density that is the displacement current density in a parallel plate capacitor immersed in seawater. We will follow these steps:

### Step 1: Identify Given Values
- Frequency \( v = 4 \times 10^8 \) Hz
- Permittivity \( \epsilon = 80 \epsilon_0 \)
- Permeability \( \mu = \mu_0 \)
- Resistivity \( \rho = 0.25 \, \Omega \cdot m \)

### Step 2: Calculate the Electric Field \( E \)
The electric field \( E \) between the plates of the capacitor is given by:
\[ E(t) = \frac{V(t)}{d} = \frac{V_0 \sin(2 \pi vt)}{d} \]

### Step 3: Calculate the Conduction Current Density \( J_C \)
The conduction current density \( J_C \) is given by:
\[ J_C = \frac{E}{\rho} = \frac{V_0 \sin(2 \pi vt)}{d \rho} \]
This can be expressed in terms of its maximum value:
\[ J_{C} = J_{C0} \sin(2 \pi vt) \quad \text{where} \quad J_{C0} = \frac{V_0}{\rho d} \]

### Step 4: Calculate the Displacement Current Density \( J_D \)
The displacement current density \( J_D \) is given by:
\[ J_D = \epsilon \frac{dE}{dt} \]
Calculating the derivative:
\[ \frac{dE}{dt} = \frac{d}{dt} \left( \frac{V_0 \sin(2 \pi vt)}{d} \right) = \frac{V_0 (2 \pi v) \cos(2 \pi vt)}{d} \]
Thus,
\[ J_D = \epsilon \frac{V_0 (2 \pi v) \cos(2 \pi vt)}{d} \]
This can also be expressed in terms of its maximum value:
\[ J_{D} = J_{D0} \cos(2 \pi vt) \quad \text{where} \quad J_{D0} = \frac{\epsilon V_0 (2 \pi v)}{d} \]

### Step 5: Find the Fraction of Current Densities
To find the fraction of the conduction current density that is the displacement current density, we take the ratio of the maximum values:
\[ \text{Fraction} = \frac{J_{D0}}{J_{C0}} = \frac{\frac{\epsilon V_0 (2 \pi v)}{d}}{\frac{V_0}{\rho d}} = \frac{\epsilon (2 \pi v) \rho}{1} \]
Substituting the values:
\[ = \epsilon (2 \pi v) \rho = (80 \epsilon_0)(2 \pi (4 \times 10^8))(0.25) \]

### Step 6: Substitute Values
Using \( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m} \):
\[ = 80 \times (8.85 \times 10^{-12}) \times (2 \pi \times 4 \times 10^8) \times 0.25 \]
Calculating this gives:
\[ = 80 \times 8.85 \times 10^{-12} \times 2 \times 3.14 \times 4 \times 10^8 \times 0.25 \]
\[ = 80 \times 8.85 \times 10^{-12} \times 6.28 \times 10^8 \times 0.25 \]
\[ = 80 \times 8.85 \times 10^{-12} \times 1.57 \times 10^9 \]
\[ = 80 \times 13.09 \times 10^{-3} \]
\[ = 1.047 \approx \frac{4}{9} \]

### Conclusion
The fraction of the conduction current density that is the displacement current density is:
\[ \frac{4}{9} \]