This question is due to many says, as found many online, it's real:

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All you need to know about capacitors is revealed by the differential form for the current.





At t = 0, the voltage across the capacitor is zero and the voltage across the resistor is V0. The initial current is then I(0) = V0/R. With this assumption, solving the differential equation yields


where τ0 = RC is the time constant of the system. As the capacitor reaches equilibrium with the source voltage, the voltages across the resistor and the current through the entire circuit decay exponentially. In the case of a discharging capacitor, the capacitor's initial voltage (VCi) replaces V0. The equations become

 
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As an analogy, a capacitor can be compared to an elastic membrane across a pipe. Pressure (voltage) causes it to stretch and store some flow / current, proportional to the pressure / voltage.

If the voltage / pressure reduces, the membrane starts to spring back to its neutral position, and forces some of the stored flow / current back.

A capacitor "membrane" is a fixed insulating layer rather than elastic and the storage is electrostatic, but they are equivalent functionally.
 
I like to think of a capacitor as a (very) small battery that can be charged and discharged millions of times per second.
I also like the water analogy but never thought of an elastic membrane, more a tank but the membrane works better.

Mike.
 
logical

Some do this for sparking within brushed DC motor contact-cut periods by connecting (two) capacitor(s) across motor terminals
 
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