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My question is with reference to this question - ESD protection using a capacitor

I am unable to grasp the relation between charge, voltage, and capacitance, although I can understand all these terms in individual context.

When they come together and get related in this ESD question, I have poor clarity.

Voltage or potential difference between any two points, is the work required to move a charge from one point to another.

However, electric charge is defined as the force that it experiences when placed in an electromagnetic field.

And capacitance, is the ability of an object to store charge.

So, when the accepted answer, states that, when something can have high voltage but low charge, how is it possible and how can I understand it intuitively?

I just want to understand this concept of charge, voltage, and capacitance more intuitively. An analogy would be better.

How is the capacitance value of our body taken as 100pF?

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The relation between charge Q, voltage V, and capacitance C is Q = V*C.

If something has high voltage, and low charge, then it has low capacitance.

The capacity of the human body with respect to ground is often quoted as 100pF. This is an approximate, ballpark, figure.

When you calculate the capacitance of two conductors in a given geometry from first principles, and you have circular or spherical symmetry, for instance a human body which looks a bit like a ball, above a ground that looks like an infinite half plane, then the capacitance tends to vary as the logarithm of the ratio of conductor diameters. That is, it varies very slowly. This means you can change the diameter of the body, or the distance above the half plane, by orders of magnitude, and the capacitance changes by only factors of two. For things about the size of a human, and typical spacings to the ground and grounded objects, the capacitance comes out within a factor of 2 or 3 to 100pF.

The way you would measure the capacitance of a human body with respect to ground is to charge it up to a given voltage, then see how much charge flows to ground when you discharge it. C=Q/V.

Just in case you're wondering ... If the capacitance of a human body with respect to ground depends somewhat on the distance above ground, then what happens when a charged human climbs onto a chair? Glad you asked. If the amount of charge on them stays the same, then as their capacitance varies, their voltage varies inversely. In fact there are electrical machines that use this constant charge variable capacitance principle to generate high voltages, the electrophorus, Wimshurst and Van Der Graff generators for instance. Wikipedia will give you details on all three.

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  • \$\begingroup\$ Thanks for the answer. my question is, when you say when something has a high voltage and low charge, it has low capacitance. How to understand this intuitively \$\endgroup\$ – Newbie Feb 9 at 12:38
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    \$\begingroup\$ Charge is the product of voltage and capacitance. If the charge is low, then at least one of the voltage and capacitance must be low, even allowing the other one to be high. Think of thimbles, or pans, or pools, or lakes of water as an analogy. The total amount of water, so volume or mass, is charge. Voltage is the depth of the water. Capacitance is the area of the container. This holds as well when you alter the area of the container and hold the charge constant, squeeze a plastic cup half-full of water, and the level rises. \$\endgroup\$ – Neil_UK Feb 9 at 14:21

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