The red plate in the above diagram is losing electrons (normally), hence it is a positive plate. Likewise, the blue capacitor plate is accepting electrons generated by the battery. Now, P.D (potential difference) is produced due to the difference in the nature of charges(Positive and negative) in the capacitor plates. More the difference in the charges (Eg. +5C and -5C), more is the potential difference available.
Now, in the diagram, you are trying to supply an electron beam to the red capacitor plate, which already contains positive charges, thus in doing so, you are trying to neutralise/cancel out the positive charges on the positive plate, hence decreasing the difference in charges in between two capacitor plates, which ultimately decreases the net potential difference in between the capacitor plates.
Now, normally, when we connect a capacitor in a circuit containing a battery, the capacitor tries to reach the source voltage of the battery, and when it does so, no current starts flowing in the circuit(when its reaches the same value as the source). In your case, the capacitor always tries to reach the source voltage of the battery, and when you supply electron beam to the positive plates of the capacitor, as i mentioned, you are trying to negate the positive charges, decreasing the p.d of the capacitors, now again the capacitor tries to reach the same voltage level of the battery, now, normally, after a long use of battery, the voltage of the battery do decrease, but in your case, at the instant you supply electron beam, it only affects the pd between the capacitor plates and not the voltage source of the battery.