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I have grounded one end of my capacitor after charging it but the voltage drops at a steady pace not as if it has lost charge. Is this because the opposing charges on the opposite plate are keeping the charges in place? When both plates are connected we have a voltage drop and V=ED and F=EQ but when one plate is grounded and the other isnt we do not have a voltage drop, why is that?

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    $\begingroup$ Because it is still an open circuit. $\endgroup$ Commented Aug 20, 2020 at 3:12

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I believe you were correct, the charges on the other plate are attracting the charges on the grounded plate keeping them from going to ground. $F=q\cdot q/r^2$, the capacitor has almost no distance separation between the 2 plates so $r$ is very small, so the charges on either end hold the opposite charges in place. I would be curious to know if the voltage drops by a little when one side is grounded or if grounding the positive vs the negative (or vise versa) side makes the voltage dissipate any faster.

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  • $\begingroup$ My capacitors dont hold charge very well and the voltage is constantly dropping even when completely disconnected (there is charge flowing through the dielectric/electrolyte) there isnt a significant difference in rate of drop between +ve and -ve but when i do flip flop on the same object there is a faster drop the 2nd time around $\endgroup$
    – ChemEng
    Commented Aug 20, 2020 at 3:17
  • $\begingroup$ That makes sense, if you hold the ground at one point some of the charges could go to ground while the majority stay held in place by the opposite charges, also as more charges go to ground, the repulsive forces on that plate decrease. But when you then move the ground over to the other side there are less charges holding them in place allowing more charges to go to ground thus dropping the voltage more significantly. $\endgroup$
    – Corey
    Commented Aug 20, 2020 at 3:23

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