To solve this problem, we need to understand the concept of displacement current.$\\$When a parallel plate capacitor is being charged, the electric field between the plates changes with time.$\\$This changing electric field gives rise to a displacement current, as introduced by James Clerk Maxwell.$\\$The displacement current $I_d$ is given by:$\\ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \\$where $\epsilon_0$ is the permittivity of free space and $\Phi_E$ is the electric flux.$\\$For a parallel plate capacitor, the electric flux $\Phi_E$ is related to the charge $Q$ on the plates:$\\ \Phi_E = \frac{Q}{\epsilon_0} \\$Thus, the rate of change of electric flux is:$\\ \frac{d\Phi_E}{dt} = \frac{1}{\epsilon_0} \frac{dQ}{dt} \\$Since the current $I$ in the circuit is $\frac{dQ}{dt},$ we have:$\\ I_d = \epsilon_0 \cdot \frac{1}{\epsilon_0} \cdot I = I \\$Therefore, the displacement current $I_d$ is equal in magnitude to the conduction current $I. \\$The direction of the displacement current is the same as that of the conduction current.$\\$Thus, the correct option is:$\\$Option 4: Displacement current of magnitude equal to $I$ flows in the same direction as $I.$