In the schematic you have shown, it is straightforward what will happen. During the periods in which the switch is closed, the DC source will transfer electrons through itself until the voltage difference between
- the left antenna element and
- all the conductors from the positive terminal to the capacitor plate, including the switch
is at the target voltage. (The right antenna element and the capacitor plate it is attached to are 'floating' — there is no way for electrons to enter or leave — so the voltage is unspecified, though even if everything starts out at equal voltage there may be a small voltage difference between the plate and the antenna due to electrostatic induction across the capacitor.)
Then, nothing further will happen, because there are no other paths for current to flow. When the switch is open, nothing happens because there are even fewer paths for current to flow.
Will this circuit radiate? Yes, a very tiny amount, when the DC source is powered on or when you start closing the switch for the first time. But it will not continue to do so, because there is no oscillation in voltage or current. You're opening and closing a switch where the two sides will be equal in voltage, and so no current flows as a consequence of the closing.