AC voltage V(t) = 20 sin(omega t) volt of frequency 50 Hz is applied to a parallel plate capacitor.

Q: Why is displacement current equal to conduction current?

By Ampere-Maxwell's law, the total current must be continuous. The conduction current stops at the plates, but the changing electric field creates an equivalent displacement current between them.

Q: Why did we use $V_0$ instead of $V_{rms}$?

The question asks for the *amplitude* (peak value) of the displacement current, so we must use the peak voltage $V_0 = 20 \text{ V}$ from the equation $V(t) = 20 \sin\omega t$.

Q: How did $100\pi$ become $25 \times 4\pi$?

It's a mathematical trick! We know the value of $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$. By factoring $100\pi$ into $25 \times 4\pi$, we can easily group $4\pi\varepsilon_0$ in the denominator and substitute the constant without dealing with messy decimals.

Q: What is the unit of $\varepsilon_0$?

The standard SI unit is Farads per meter (F/m), which is equivalent to $C^2 N^{-1} m^{-2}$.

Q: Does the displacement current depend on the frequency?

Yes! Higher frequency means a faster changing electric field, which leads to a higher displacement current. Mathematically, $I_0 = V_0 \omega C$, so it is directly proportional to $\omega$.

Question

AC voltage $V(t) = 20 \sin\omega t$ volt of frequency $50 	ext{ Hz}$ is applied to a parallel plate capacitor. The separation between the plates is $2 	ext{ mm}$ and the area is $1 	ext{ m}^2$. The amplitude of the oscillating displacement current for the applied AC voltage is …… .
(Take, $\varepsilon_0 = 8.85 	imes 10^{-12} 	ext{ F/m}$)

Options:

(A) $21.14 \mu	ext{A}$

(B) $83.37 \mu	ext{A}$

(C) $27.79 \mu	ext{A}$

(D) $55.58 \mu	ext{A}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a simple yet fascinating circuit: an alternating voltage source connected to a parallel plate capacitor. The voltage is given by the equation $V(t) = 20 \sin\omega t$, which tells us that the peak voltage $V_0$ is $20 	ext{ V}$. The frequency of this AC source is $50 	ext{ Hz}$.

The capacitor itself has plates with a generous area of $A = 1 	ext{ m}^2$, separated by a tiny gap of $d = 2 	ext{ mm}$, or $2 	imes 10^{-3} 	ext{ m}$. Our goal is to find the amplitude (the peak value) of the oscillating displacement current between these plates.

### The Master Equation

To find the displacement current, we must remember a beautiful consequence of Maxwell's equations: the displacement current $I_d$ flowing through the gap between the capacitor plates is exactly equal to the conduction current $I_c$ flowing through the connecting wires.

Therefore, the amplitude of the displacement current, $I_0$, is simply the peak voltage divided by the capacitive reactance $X_C$:

$$I_0 = \frac{V_0}{X_C}$$

We know that the capacitive reactance is given by $X_C = \frac{1}{\omega C}$. Let's first set up the expression for the capacitance $C$ using the physical dimensions of the capacitor:

$$C = \frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 	imes 1}{2 	imes 10^{-3}}$$

We also need the angular frequency $\omega$. Since the frequency $f$ is $50 	ext{ Hz}$, we have:

$$\omega = 2\pi f = 2\pi(50) = 100\pi 	ext{ rad/s}$$

### A Mathematical Trick

Now, let's plug this capacitance into the reactance formula. Notice how we keep $\varepsilon_0$ as a symbol for now. This is a smart move because $\frac{1}{4\pi \varepsilon_0}$ is a standard constant ($9 	imes 10^9$) that we can exploit.

$$X_C = \frac{1}{100\pi \left(\frac{\varepsilon_0}{2 	imes 10^{-3}}ight)} = \frac{2 	imes 10^{-3}}{100\pi \varepsilon_0}$$

Here is the catch. We can rewrite $100\pi$ as $25 	imes 4\pi$. This allows us to isolate the $\frac{1}{4\pi \varepsilon_0}$ term:

$$X_C = \frac{2 	imes 10^{-3}}{25 	imes 4\pi \varepsilon_0} = \frac{2 	imes 10^{-3}}{25} 	imes \left(\frac{1}{4\pi \varepsilon_0}ight)$$

Substituting $\frac{1}{4\pi \varepsilon_0} = 9 	imes 10^9$, the reactance becomes:

$$X_C = \frac{2 	imes 10^{-3}}{25} 	imes 9 	imes 10^9 = \frac{18}{25} 	imes 10^6 \, \Omega$$

### Final Calculation

Let's substitute the values and get the final answer. The peak voltage is $20 	ext{ V}$. Dividing this by our calculated reactance, we get:

$$I_0 = \frac{20}{\frac{18}{25} 	imes 10^6} = \frac{20 	imes 25}{18} 	imes 10^{-6} 	ext{ A}$$

$$I_0 = \frac{500}{18} \mu	ext{A} \approx 27.78 \mu	ext{A}$$

So, the amplitude of the displacement current is approximately **$27.79 \mu	ext{A}$**, which perfectly matches option (c).

This problem beautifully demonstrates how a changing electric field across a vacuum gap acts exactly like a physical current in a wire, maintaining the continuity of the circuit!

Select Answer:

Visualized Solution

Circuit Setup

Displacement Current Amplitude

Capacitance Calculation

Capacitive Reactance

Simplifying Reactance

Peak Current

Final Calculation

Conceptual Takeaway

The Sigma Insight: Displacement Current

Analyzing the Setup

The Master Equation

A Mathematical Trick

Final Calculation

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