In the AC circuit shown in the figure, a pure capacitor of...

For a pure capacitor, the relationship between the instantaneous current $I$ and the charge $q$ is given by:

$I = \dfrac{dq}{dt}$

Since the charge on the capacitor is $q = C V_C$, we have:

$I = C \dfrac{dV_C}{dt}$

Rearranging the equation to solve for $V_C$ gives:

$dV_C = \dfrac{I}{C} dt$

Integrating both sides with respect to time $t$:

$V_C = \int \dfrac{I}{C} dt$

Given that the instantaneous current is $I = I_0 \cos \omega t$, substituting this into the integral yields:

$V_C = \int \dfrac{I_0 \cos \omega t}{C} dt$

$V_C = \dfrac{I_0}{C} \int \cos \omega t dt$

$V_C = \dfrac{I_0}{C} \left( \dfrac{\sin \omega t}{\omega} \right)$

$V_C = \dfrac{I_0}{\omega C} \sin \omega t$

Answer: $\dfrac{I_0}{\omega C} \sin \omega t$