AC divider circuit used to charge a capacitor
But can you explain using charge flow why this waveform is occuring at
Vx. Just for understanding only. Thanks – Hyde

Simplify the circuit: -

I've combined (with absolutely no loss of functionality or performance) the three 45 pF capacitors into one 135 pF capacitor. To do this I must also reduce the input voltage of this equivalent circuit from 311 volts pk to 104 volts pk.
Can you see why trying to charge the 1 uF capacitor from the source via a capacitor that is 7,407 times lower in value will take some time and doesn't happen instantly?

If you connected a diode to Vx and ground (anode to ground), it forces Vx to stay at a positive value for much longer in the AC cycle and charges the output capacitor more strongly. In effect, the added diode allows conduction through the capacitor (135 pF) when the AC signal goes negative as well as when the AC signal is positive (current through the existing D1 diode).