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

The correct answer is:
$I_d=V_0\omega Ccos\omega t$

Step-by-Step Explanation:

1. Understanding the relationship between Voltage and Charge:
We are given a capacitor of capacitance C connected across an AC voltage source. The instantaneous voltage across the plates of the capacitor is given by:
$V=V_{0}sin\omega t$
The instantaneous charge q on the plates of the capacitor is related to the instantaneous voltage V by the formula:
$q=CV$
Substituting the expression for V into the charge equation, we get:
$q=C{V}_{0}sin\omega t$

2. Displacement Current in a Capacitor:
According to Maxwell's equation, the displacement current ($I_d$) between the plates of a capacitor is equal to the rate of change of electric flux, which is physically equivalent to the conduction current ($I_c$) flowing through the connecting wires. Thus, we can find the displacement current by differentiating the charge q with respect to time t:
$I_d=\frac{dq}{dt}$

3. Performing the Differentiation:
Now, we differentiate the expression for q with respect to t:
$I_d=\frac{d}{dt}(C{V}_{0}sin\omega t)$
Since C and $V_0$ are constants, they can be pulled out of the derivative:
$I_d=C{V}_0\frac{d}{dt}(sin\omega t)$
Using the chain rule of differentiation, we know that $\frac{d}{dt}(sin\omega t)=\omega cos\omega t$. Therefore, we obtain:
$I_d=V_0\omega Ccos\omega t$
This matches the exact expression shown in the correct option image (image_0.png).