A capacitor has been charged by a dc source. What are the magnitude of conduction and displacement current, when it is fully charged?

Hint :The conduction current in a capacitor refers to the flow of electrons in the capacitor plates.
And, the displacement current is due to the electrons that are displaced in the time varying electric field, between the plates.

Complete Step By Step Answer:
A capacitor has 2 conductor plates (the positive plate, $ Q + $ and the negative
plate, $ Q - $ ) separated by a dielectric medium (air) .
Suppose a capacitor of capacitance C is charged by a D.C source (a battery of 5 V) , it will gradually build up charge on each plate ( positive charge on $ Q + $ and equal , negative charge on $ Q - $ ) . A capacitor stores energy in the form of an electrostatic field between its plates. Now, current flows through the capacitor by time varying electric fields.
Therefore, the actual charge Q on the plates of the capacitor is calculated as
 $ Q = CXV $ .
Where, $ V $ denotes the Voltage, in Volts
 $ \Rightarrow C = dfrac{Q}{V} $
The rate of flow of charge with respect to time is termed as current.
Conduction Current $ {i_c} $ is given by $ {i_c} = Cdfrac{{dV}}{{dt}} $
After a certain period of time, the potential difference on the plate is equal to the maximum applied voltage (5 V) .Since the voltage is constant, the charge will become constant and therefore the conduction current will be zero $ \left( {\because dfrac{{dV}}{{dt}} = 0} \right) $ .
 $ \Rightarrow {i_c} = 0 $
Since there is no change in voltage, the electric field is not time varying . There will be no movement of charge and no displacement current, which is given as
 $ \Rightarrow {i_d} = 0 $
 $ \Rightarrow {i_c} = {i_d} = 0 $
 $ \therefore $ Conduction current $ = 0 $
 $ \therefore $ Displacement current $ = 0 $ .

Note :
However, when an ac source is used and the capacitor is fully charged, the current flows continuously. The capacitor circuit does not act as an open circuit and does not block the current flow.