A capacitor of capacitance ‘C’, is connected across an ac source of voltage V, given by V=V₀ sinωt The displacement current between the plates of the capacitor, would then be given by :

The correct option is I_d = V₀ωCcosωt.

Step-by-Step Explanation:

1. Understanding Capacitor and Voltage:
We are given a capacitor of capacitance $C$ connected to an alternating voltage source. The voltage across the capacitor as a function of time $t$ is given by:
$V=V_{0}sin\left(\omega t\right)$

2. Relation between Charge and Voltage:
The instantaneous charge $q$ on the plates of the capacitor at any time $t$ is related to the capacitance and the instantaneous voltage across it by the relation:
$q=CV$

Substituting the expression for $V$ into the equation:
$q=C{V}_{0}sin\left(\omega t\right)$

3. Calculating the Displacement Current:
By definition, the displacement current $I_d$ between the plates of a capacitor is equal to the rate of change of charge on its plates, which is also equal to the conduction current in the wires. Mathematically:
$I_d=\frac{dq}{dt}$

Substitute the expression for $q$:
$I_d=\frac{d}{dt}[C{V}_{0}sin(\omega t\left)\right]$

Since $C$ and $V_0$ are constants, we can take them out of the derivative:
$I_d=C{V}_0\frac{d}{dt}[sin(\omega t\left)\right]$

4. Evaluating the Derivative:
Using the chain rule of differentiation, the derivative of $sin\left(\omega t\right)$ with respect to $t$ is $\omega cos\left(\omega t\right)$:
$I_d=C{V}_0(\omega cos(\omega t\left)\right)$

Rearranging the terms, we get:
$I_d=V_0\omega Ccos\left(\omega t\right)$

Thus, the displacement current between the plates of the capacitor is given by $I_d=V_0\omega Ccos\left(\omega t\right)$.