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I want to learn about this way of charging the capacitor. At my university, we charge capacitor with power supply. Its negative power supply. Power supply is grounded (earthed). A conductor from power supply is attached to one plate of capacitor and other plate of capacitor is grounded (earthed) separately. Both earthed points are different (physically). I want to learn how this capacitor is getting charged?

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Both earthed points are different (physically). I want to learn how this capacitor is getting charged

The fact that the power supply and one plate of the capacitor are earth grounded at different locations simply potentially introduces additional resistance through which charging occurs. That resistance increases the charging time constant (t=RC) slowing down the rate of charging the capacitor. How slow for a given capacitance C depends on how much resistance exists between the earth connection of the power supply and the earth connection of the one plate, which in turn depends on the conductivity of the soil and the distance between the connections.

Perhaps the diagram below illustrates your university experiment. It assumes there is no other resistor connected between the power supply and the capacitor, i.e., the only resistance is the soil between the earth connections and the resistance of the connecting wires is negligible. Under these conditions if the capacitor is initially uncharged, and the switch shown closes at time $t=0$, the theoretical voltage across the capacitor as a function of time, $V_{C}(t)$, would be

$$V_{C}(t)= V(1-e^{-t/RC})$$

If your experiment involved another resistor connected between the power supply and capacitor, you would need to add it to the value of $R$ in the above equation.

I want to add a further question. What would happen if we remove the supply and connect a ground stick (connected to another separate ground) at the positive terminal of the capacitor shown in the figure above. How would a charged capacitor behave in this scenario?

The capacitor will discharge through the ground. In this case the voltage across the capacitor as a function of time will be $$V_{C}(t)=V_{o}\epsilon ^{-t/RC}$$ where $V_{o}$ is the initial voltage across the capacitor.

Hope this helps.

enter image description here

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  • $\begingroup$ Thanks alot Bob D. I want to add a further question. What would happen if we remove the supply and connect a ground stick (connected to another separate ground) at the positive terminal of the capacitor shown in the figure above. How would a charged capacitor behave in this scenario? $\endgroup$ Oct 22 '21 at 19:44
  • $\begingroup$ Are you asking what would happen if you started with a charged capacitor and connected both the positive and negative plates to earth with separate grounding electrodes? $\endgroup$
    – Bob D
    Oct 22 '21 at 20:02
  • $\begingroup$ Exactly. I am asking the same. $\endgroup$ Oct 23 '21 at 7:09
  • $\begingroup$ @MuhammadMuzammilNawaz I have updated my answer to respond to your additional question. $\endgroup$
    – Bob D
    Oct 23 '21 at 19:03
  • $\begingroup$ @BobD it would discharge even faster, as the earth grounding acts as an infinite sink for charge, it does not path the discharge only through the measured soil resistance R between the contacts. $\endgroup$
    – PcMan
    Oct 23 '21 at 19:06

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