Derivation of energy stored in a capacitor

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Trying to understand the derivation of energy stored in a capacitor:

The energy (measured in Joules) stored in a capacitor is equal to the work done to charge it. Consider a capacitance C, holding a charge +q on one plate and -q on the other. Moving a small element of charge dq from one plate to the other against the potential difference V = q/C requires the work dW:

dW = q/ C dq

where

W is the work measured in joules

q is the charge measured in coulombs

C is the capacitance, measured in farads


The dq doesnt seem to fit for me. It just seems as if someone put it there for the derivation to work and their is no basis for it being there.
 

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  • #2
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Are you used to the formula of the energy stored in an electirc field

[tex]W = \frac{\varepsilon_0}{2} \int \limits_\mathcal{V} \mathrm dr^3 ~ \vec E^{\, 2}(\vec r)[/tex]​

?
 
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  • #3
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Thats a bit complex.
Im trying to understand the derivation for basic equation of energy stored in a capacitor.
E = 0.5 CV^2
or E = 0.5 Q^2/C
 
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  • #4
Doc Al
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The dq doesnt seem to fit for me. It just seems as if someone put it there for the derivation to work and their is no basis for it being there.
Do you understand how to find the work required to move a charge through a potential difference? Since the voltage changes as the charge accumulates you need to look at the increment of work, dW = Vdq. (If the voltage were constant, the work would simply equal VQ.)
 
  • #5
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Maybe you know that the electric field is crucial for your consideration!?

I know that's slightly abstract, but the energy is not stored in the capacitor itself. It is stored in the electric field (only the sources and sinks of the electric field are located on the capacitor's plates).

In my opinion all other derivations want to bypass this abstract point of view.


But you have luck, that the equation i have stated is easy to compute in the case you have mentioned!
You know that in the plate capacitor there is a uniform electric field (doesn't depend on the position within the plates), so we can write the expression as


[tex]W = \frac{\varepsilon_0}{2} \, \vec E^2 \, \int \limits_\mathcal{V} ~ \mathrm dr^3[/tex]​

(by the way, i think W is a much better designation for the energy than E, i'm sorry for that).
The volume of the plate capacitor is


[tex]V = \int \limits_\mathcal{V} ~ \mathrm dr^3 = A \cdot d[/tex]​

and the magnitude of the electric field is


[tex]|\vec E| = \frac{V}{d}[/tex]​

this applied to the formula stated above yields


[tex]W = \frac{\varepsilon_0}{2} \, \Bigl(\frac{V}{d}\Bigr)^2 \, A \cdot d = \frac{1}{2} \, V^2 ~ \varepsilon_0 \cdot \frac{A}{d} = \frac{1}{2} V^2 C[/tex]​

The last term is actually valid for all capacitors of any geometry.

I hope, i've teased you!?

with best regars
 

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