Animated Solution for Physics - Electromagnetic Waves: Seawater at a frequency $f=9\times 10^2$ Hz, has permittivity $ϵ=80{ϵ}_0$ and resistivity $r=0.25\Omega \text{-m}$. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source $V(t)=V_0\sin (2\pi ft)$. Then, the conduction current density becomes $10^x$ times the displacement current density after time $t=\frac{1}{800}$ s. The value of $x$ is ........... . $\left(\text{Take, }\frac{1}{4\pi {ϵ}_0}=9\times 10^9{\text{N-m}}^2{\text{C}}^{-2}\right)$

A Tale of Two Currents

Imagine a parallel plate capacitor completely submerged in seawater. Now, seawater isn't just a passive dielectric; it's packed with free ions like sodium and chloride, making it a decent conductor too! This means we are dealing with a leaky capacitor. When we apply an alternating voltage, two distinct types of currents will flow simultaneously between the plates: the conduction current (due to the physical movement of ions) and the displacement current (due to the changing electric field polarizing the water molecules).

Let's first talk about the conduction current. Remember the macroscopic form of Ohm's Law? We know that $I_c=\frac{V(t)}{R}$.

To find the current density $J_c$, we divide the current by the area $A$. We can also express the resistance $R$ in terms of the seawater's resistivity $r$, the distance between the plates $d$, and the area $A$ as $R=\frac{r\cdot d}{A}$.

Substituting this into our density equation, we get:

Now let's move to the displacement current. Recall Maxwell's brilliant addition to Ampere's Law. The displacement current $I_d$ is the rate of change of the electric flux, which for a capacitor simplifies to the rate of change of charge, $I_d=\frac{d(CV)}{dt}$.

Dividing by area to get the density $J_d$, and substituting the capacitance $C=\frac{ϵA}{d}$, we find:

Differentiating our voltage function $V(t)=V_0\sin (2\pi ft)$ with respect to time yields:

The problem asks for the ratio of these two current densities at a specific time. Let's divide $J_c$ by $J_d$. Notice how the voltage amplitude $V_0$ and the plate separation $d$ cancel out beautifully!

Now let's substitute the given values. Don't make a silly mistake here with the angles. We are given $t=\frac{1}{800}$ s and $f=900$ Hz.

The phase angle becomes $2\pi ft=2\pi (900)\left(\frac{1}{800}\right)=\frac{9\pi }{4}$. The tangent of $\frac{9\pi }{4}$ (which is $2\pi +\frac{\pi }{4}$) is simply $1$. So the trigonometric part simplifies perfectly.

Next, let's evaluate the constant multiplier. We are given $r=0.25\Omega \text{-m}$ and $ϵ=80{ϵ}_0$. To make things easy, we use the given value $\frac{1}{4\pi {ϵ}_0}=9\times 10^9$, which means ${ϵ}_0=\frac{1}{36\pi \times 10^9}$.

After careful calculation, the entire denominator beautifully collapses:

Substituting everything back into our ratio equation, we get:

The question states this ratio is $10^x$. Comparing the exponents, we arrive at our final answer:

At this specific frequency, the conduction current is a million times stronger than the displacement current. However, notice that $J_d$ is proportional to the frequency $f$. If we were to crank up the frequency to the gigahertz range, the displacement current would completely dominate, and the seawater would behave almost like a perfect dielectric!