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Based on the research I did online and after reading basic electrical books I understand that a capacitor stores energy in the form of electric fields. So I understand that, to charge a capacitor, all you need is electric field, you do not need current to charge a capacitor, is that correct?

If we can charge a capacitor with only electric fields where little or no current flows, can we charge a capacitor of required rating with high voltage electric fields (but no or little current) coming out of the secondary side of a step up transformer?

The step up operation steps up the voltage to very high levels where we can see arcs if both terminals of secondary coils are brought together in close proximity. The arc shows that there's very strong electric fields between the terminals that they can easily ionises the air and discharge themselves.

If all this is possible I would love to have the formula for calculation, but more than the formulae I want to understand the concept of charging and discharging a capacitor.

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  • \$\begingroup\$ A transformer means you're dealing with AC, and you need DC to significantly charge the capacitor. \$\endgroup\$
    – Perry Webb
    Oct 25, 2022 at 9:07
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    \$\begingroup\$ @winny: That's the formula for a coil: \$U = L\frac{dI}{dt}\$. Swap U and I for a capacitor: \$I = C\frac{dU}{dt}\$. \$\endgroup\$
    – Dave Tweed
    Oct 25, 2022 at 10:41
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    \$\begingroup\$ You have to supply charge to the capacitor in order for it to store energy in an electric field. You can't use an electric field to charge a capacitor unless you can get the field to make some charge flow into the capacitor, and flowing charge is current, so ... \$\endgroup\$
    – brhans
    Oct 25, 2022 at 12:04
  • \$\begingroup\$ @DaveTweed Typed too fast once again. Sorry about that. How do you get LaTeX in comments? \$\endgroup\$
    – winny
    Oct 25, 2022 at 13:27
  • \$\begingroup\$ @winny: Same as anywhere else: \$ ... \$ for inline, $$ ... $$ for block. It's kind of weird that that works, but simple HTML entities (e.g., ±) does not. \$\endgroup\$
    – Dave Tweed
    Oct 25, 2022 at 14:30

2 Answers 2

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Based on the research I did online and after reading basic electrical books I understand that a capacitor stores energy in the form of electric fields. So I understand that, to charge a capacitor, all you need is electric field, you do not need current to charge a capacitor, is that correct?

No. You must supply charge, to charge a capacitor. A flow of charge is current. With a small capacitor, it would be a small charge, but a finite current has to flow for a finite time to charge any capacitor.

If we can charge a capacitor with only electric fields where little or no current flows, can we charge a capacitor of required rating with high voltage electric fields (but no or little current) coming out of the secondary side of a step up transformer?

As a small capacitor can be charged with little current, you can use the secondary side of a step up transformer to supply that current. Bear in mind that the output of a transformer is AC, so will repeated charge and discharge a capacitor connected to its output terminals. If you want to charge the capacitor to a DC voltage, you need a rectifier between transformer and capacitor.

The step up operation steps up the voltage to very high levels where we can see arcs if both terminals of secondary coils are brought together in close proximity. The arc shows that there's very strong electric fields between the terminals that they can easily ionises the air ...

Yes

... and discharge themselves.

The terminals are still connected to the transformer, so still being supplied with current from it.

If all this is possible I would love to have the formula for calculation, but more than the formulae I want to understand the concept of charging and discharging a capacitor.

A capacitor stores charge. We can measure the charge in Coulombs, or Ampere.seconds, and we tend to give it the symbol Q. Q = I.t, the product of the current that flows, and the time that it flows, to charge the capacitor.

We also define a capacitance for the capacitor, C = Q/V, the charge stored per volt across the capacitor.

Then we can manipulate those formulae to give other useful results. For instance, how fast does the voltage across a capacitor rise when supplied with a given current? dV/dt = I/C.

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  • \$\begingroup\$ Note: charge does not build up in a capacitor, as charge flows into one side or out of the other. When we talk about the charge in a capacitor, we're talking about the integral of current. \$\endgroup\$
    – user253751
    Oct 25, 2022 at 20:02
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So I understand that, to charge a capacitor, all you need is electric field, you do not need current to charge a capacitor, is that correct?

No, that is incorrect, you need the flow of charge (aka current) to create an electric field in a capacitor as embodied in these equations: -

$$Q = C\cdot V$$

If you convert charge to current by differentiation: -

$$I = C\cdot\dfrac{dv}{dt}$$

This tells you that if you supply a constant current (\$I\$) into a capacitor, the capacitor plates will develop a rising/falling voltage across them that changes at the rate of \$I/C\$.

The remainder of your question is falsely based on the assumption that a capacitor can be charged from an electric field with no current flow hence, it is ignored.

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