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Step-by-Step Solution
Step 1: Identify the key physical quantities
• Frequency of the AC source: f = 9 \times 10^2 \text{ Hz}
• Permittivity of sea water: \varepsilon = 80 \varepsilon_0
• Resistivity of sea water: \rho = 0.25 \,\Omega \text{m}
• Time at which the ratio of conduction to displacement current is evaluated: t = \frac{1}{800}\,\text{s}
Step 2: Express conduction and displacement current densities
1. Conduction current density ( J_c ) in a medium of conductivity \sigma and electric field E(t) is
J_c(t) = \sigma \, E(t)
where \sigma = \frac{1}{\rho} is the conductivity.
2. Displacement current density ( J_d ) for a time-varying electric field E(t) is
J_d(t) = \varepsilon \, \frac{\partial E(t)}{\partial t}\,.
Step 3: Relate the electric field to the applied AC voltage
If the capacitor is driven by V(t) = V_0 \sin\bigl(2\pi f\, t\bigr) , then the electric field can be written (up to a constant factor depending on geometry) as
E(t) = E_0 \sin(\omega t), \quad \text{where } \omega = 2\pi f.
Thus,
\displaystyle E(t) = E_0 \sin(\omega t)\quad\text{and}\quad \frac{\partial E(t)}{\partial t} = \omega\, E_0 \cos(\omega t).
Step 4: Write down J_c(t) and J_d(t) explicitly
• \sigma = \frac{1}{\rho} = \frac{1}{0.25} = 4\,(\text{S/m}).
• J_c(t) = \sigma\, E_0 \sin(\omega t) = 4 \, E_0 \sin(\omega t).
• J_d(t) = \varepsilon \, \omega \, E_0 \cos(\omega t).
Here, \varepsilon = 80 \varepsilon_0.
Step 5: Compute the ratio \frac{J_c(t)}{J_d(t)} at t = \frac{1}{800} s
The angular frequency is
\omega = 2\pi f = 2\pi \times 9 \times 10^2 = 1800\pi.
First, calculate the argument \omega t at t = \frac{1}{800} s:
\omega \left(\frac{1}{800}\right) = 1800\pi \times \frac{1}{800} = \frac{1800}{800}\,\pi = \frac{9}{4}\,\pi.
Now,
\sin\Bigl(\tfrac{9\pi}{4}\Bigr) = \frac{\sqrt{2}}{2}, \quad \cos\Bigl(\tfrac{9\pi}{4}\Bigr) = \frac{\sqrt{2}}{2}.
Hence, at t = \frac{1}{800} s, \sin(\omega t) and \cos(\omega t) have the same value. So,
\displaystyle \frac{J_c(t)}{J_d(t)}\Bigl\rvert_{t = 1/800} = \frac{\sigma\, E_0 \sin(\omega t)}{\varepsilon \, \omega \, E_0 \cos(\omega t)} = \frac{\sigma}{\varepsilon \,\omega} \times \frac{\sin(\omega t)}{\cos(\omega t)}.
Since \sin(\omega t) = \cos(\omega t) at that instant,
\displaystyle \frac{J_c(t)}{J_d(t)}\Bigl\rvert_{t = 1/800} = \frac{\sigma}{\varepsilon \,\omega}.
Step 6: Evaluate the numerical value
1. \sigma = 4\,(\text{S/m}).
2. \varepsilon = 80\, \varepsilon_0 = 80 \times 8.854 \times 10^{-12} \approx 7.0832 \times 10^{-10}\, \text{F/m}.
3. \omega = 1800\pi \approx 5655.
Thus,
\displaystyle \frac{\sigma}{\varepsilon \,\omega} \;=\; \frac{4}{(7.0832 \times 10^{-10}) \times 5655}.
Compute the denominator approximately:
(7.0832 \times 10^{-10}) \times 5655 \approx 4.0 \times 10^{-6}.
Therefore,
\displaystyle \frac{4}{4.0 \times 10^{-6}} = 10^{6}.
Hence,
\displaystyle \frac{J_c(t)}{J_d(t)}\Bigl\rvert_{t = 1/800} = 10^{6}.
So the conduction current density becomes 10^6 times the displacement current density. Thus,
\displaystyle x = 6.
Final Answer:
The required value of x is 6.
