# Solved: Verify that the conduction current in the wire equals the displacement current ...

## Question:

Verify that the conduction current in the wire equals the displacement current between the plates of the parallel plate capacitor in the figure below.

The voltage source supplies $V_c = V_o\sin\omega {t}$.

## Solution:

The question is asking us to compare if the conduction current $I_c$ and the displacement current $I_d$ are equal.

Thus, we need to calculate for each.

### 1. Conduction current (Ic) = ?

From the circuit, we can see that, the current flowing in the circuit is the same as the current passing through the capacitor.

Hence, we can use $I_c = C\frac{\text{d}V_c}{\text{d}t}$.

Where:

$C$ is capacitance of the capacitor. $C =\frac{ \epsilon A}{d}$

$V_c$ is the voltage across the capacitor, which is the same as the input voltage source $V_c = V_o\sin\omega{t}$.

Therefore, $I_c = C\frac{d(V_o\sin\omega t)}{dt}$

If we differentiate the voltage, we have $I_c = C V_o \omega\cos\omega t$.

Now letโs find the displacement current $I_d$.

### 2. Displacement current $(I_d)$ = ?

But $I_d = \int\int J_d \cdot ds$ (surface integral of displacement current density over a surface)

This is also equal to the surface integral of $\int\int\frac{\partial D}{\partial t} \cdot ds$

Where:

$D$ (is electric flux density). It can also be expressed in terms of electric field intensity ($E$) as $D = \epsilon E$

Hence, $I_d = \int\int J_d \cdot ds = \int\int\frac{\partial D}{\partial t} \cdot ds$.

But $E$ can also be expressed in terms of $\frac{V }{D} \frac{(voltage [V])}{(distance [m])}$

Hence, $D = \epsilon\frac{V }{D} \frac{(voltage [V])}{(distance [m])}$

So to solve for $I_d = \int\int\frac{\partial D}{\partial t} \cdot ds$, we need to differentiate $D$.

Therefore, $I_d = \int\int\frac{\partial D}{\partial t} \cdot ds$.

$\frac{\partial D}{\partial t} = \frac{\epsilon\cdot V_o}{d} \omega\cos\omega t$

$I_d = \int\int(\frac{\epsilon\cdot V_o}{d} \omega\cos\omega t) \cdot ds$

$= \int\int ds (\frac{\epsilon\cdot V_o}{d} \omega\cos\omega t)$.

$\int\int ds$ = Area of the surface $(A)$

By substituting $A$, $I_d = A (\frac{\epsilon\cdot V_o}{d} \omega\cos\omega t)$

Remember that $C = \frac{\epsilon A}{d}$

Therefore, $I_d = C V_o\omega\cos\omega t$.

So in conclusion, by comparing $I_c$ and $I_d$, we can see that they are equal. i.e., $I_c = I_d = C V_o\omega\cos\omega t$.