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 answer is:
$I\_d\; =\; V\_0\; \backslash omega\; C\; \backslash cos\; \backslash omega\; t$

Step-by-step derivation:

1. Recall the definition of displacement current ($I\_d$):
Displacement current is defined in terms of the rate of change of electric flux ($\backslash Phi\_E$) between the plates of the capacitor:
$I\_d\; =\; \backslash epsilon\_0\; \backslash frac\{d\backslash Phi\_E\}\{dt\}$

2. Express electric flux in terms of voltage:
For a parallel-plate capacitor with plate area $A$ and separation distance $d$, the electric field $E$ between the plates is given by:
$E\; =\; \backslash frac\{V\}\{d\}$
The electric flux through the region between the plates is:
$\backslash Phi\_E\; =\; E\; \backslash cdot\; A\; =\; \backslash frac\{V\; \backslash cdot\; A\}\{d\}$

3. Relate flux to capacitance:
The capacitance $C$ of a parallel-plate capacitor is given by:
$C\; =\; \backslash frac\{\backslash epsilon\_0\; A\}\{d\}$
Rearranging this gives:
$\backslash frac\{A\}\{d\}\; =\; \backslash frac\{C\}\{\backslash epsilon\_0\}$
Substitute this back into the electric flux expression:
$\backslash Phi\_E\; =\; V\; \backslash left(\backslash frac\{C\}\{\backslash epsilon\_0\}\backslash right)$

4. Substitute flux into the displacement current formula:
$I\_d\; =\; \backslash epsilon\_0\; \backslash frac\{d\}\{dt\}\; \backslash left(\; \backslash frac\{V\; C\}\{\backslash epsilon\_0\}\; \backslash right)\; =\; C\; \backslash frac\{dV\}\{dt\}$

5. Calculate the derivative of the AC voltage source:
The voltage across the AC source is given by:
$V\; =\; V\_0\; \backslash sin\; \backslash omega\; t$
Differentiating $V$ with respect to time $t$:
$\backslash frac\{dV\}\{dt\}\; =\; \backslash frac\{d\}\{dt\}\; (V\_0\; \backslash sin\; \backslash omega\; t)\; =\; V\_0\; \backslash omega\; \backslash cos\; \backslash omega\; t$

6. Find the final displacement current:
Substitute the derivative $\backslash frac\{dV\}\{dt\}$ into the equation for $I\_d$:
$I\_d\; =\; C\; (V\_0\; \backslash omega\; \backslash cos\; \backslash omega\; t)\; =\; V\_0\; \backslash omega\; C\; \backslash cos\; \backslash omega\; t$