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So if use these two equations:

$$J=C\times V$$

and:

$$F=\frac{C}{V}$$

I get:

$$J=F\times V^2$$

Then why is it divided by two? Is it always like this or only for capacitors?

Tom Carpenter
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Andrés
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    Do you know calculus? It's the result of an integral. Similar to the kinetic energy equation in physics. – MarkU Mar 03 '17 at 21:54
  • @MarkU The result of what integral? – Andrés Mar 03 '17 at 22:01
  • http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html – Claudio Avi Chami Mar 03 '17 at 22:04
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    $ {1 \over 2} (\cdot) (\cdot) ^2$ is very common! This is the result of $ (\cdot) \int (\cdot) d(\cdot) $. – skvery Mar 03 '17 at 22:13
  • @Claudio Avi Chami Thanks a lot Claudio. But why is it not in concordance with the other two formulas? Thanks. – Andrés Mar 03 '17 at 22:17
  • This question has been asked on the physics stack: Why does a 1/21/2 appear in this simple capacitor equation? – The Photon Mar 03 '17 at 22:24
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    @Andrés: you have an error in your thoughts. Energy is the integral of power over time. But power isn't constant over time when loading a capacitor, and so you don't have a rectangle of V * V, but a triangle of 1/2 * V * V. You cannot solve this without either applying calculus or wrapping your mind around the triangle (which is what I learned in 9th grade high school physics, before learning calculus.) – Janka Mar 03 '17 at 22:26
  • @The Photon Thanks a lot, i get it now. It was driving me crazy, love ya :) – Andrés Mar 03 '17 at 22:28
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    @Janka That actually makes a lot of sense. Im starting calculus next year at collegue but i think i get that. Thanks for your help :) – Andrés Mar 03 '17 at 22:33
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    In calculus "W" means energy, it stands for "Work", equations don't always use symbols that match their units e.g. I = current, L = inductance, C = capacitance and Q is charge despite these values having units of Amps, Henries, Farads and Coulombs respectively. The reason the equations don't give you the right answer is because you have multiplied them instead of taking an integral. – Sam Mar 03 '17 at 22:46

1 Answers1

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I'll copy my answer from Physics.SE over here, since the question is on topic for both sites:

You can calculate the energy by assuming you charge the capacitor with a constant current.

Then the energy input to the capacitor is

$$\int_0^t{IV(t')\mathrm{d}t'}$$

Since I is constant, you know that (for a linear capacitor) V(t) is a ramp function. So from geometry (the area of a triangle formula) the integral is

$$\frac{1}{2}ItV(t)$$

Now, \$It\$ is the total charge that's been given to the capacitor and \$V(t)\$ is the final voltage after charging, so this is the same as the \$\frac{1}{2}QV\$ formula you are familiar with.

Physical explanation

Imagine you start with an uncharged capacitor. The first tiny element of charge you drive onto the plates takes practically no energy, because the voltage starts at 0. As you add more charge to the plates the voltage increases and it takes more energy to add each additional element of charge. Finally, the last element takes \$V \cdot dQ.\$ energy to push onto the plates.

If all the charge took equal energy to push onto the plates, you would end up with your formula, \$E=QV\;.\$ But since the first bits of charge you put on took less energy, you end up with an overall average of just \$E=\frac{1}{2}QV\;.\$ To figure out why the pre-factor is exactly 1/2, you could either go to the integration formula, or use the geometric argument which is how I mentally calculated the integral anyway.

The Photon
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