Proof of Energy Density in Capacitor?

[onurbeyaz]

Hi, I just registered and don't know if I opened the post in right place, this is not a homework question, only something I wonder.

I know that electric field energy density in a parallel-plate capacitor is;

u= ε * (E^2) / 2

and I know how this formula derivated for parallel-plate capacitor. But I learned that this formula can be generated for every geometrical type of capacitors. I tried to derivate this formula for cylindrical capacitors but I could not make it. Are there any general proof of this formula that can be used for every geometrical type of capacitors? İf there isn't, how can this formula generated for cylindrical capacitors? Thanks for your answers

 

[Simon Bridge]

Welcome to PF;
I cannot tell how you are thinking about this so I don't know how best to reply.
The equation follows from the definition of energy density - which follows from the energy stored in an arbitrary electric field.
http://www.ece.umd.edu/class/enee380-1.F2004/lectures/lecture12.htm
... for a parallel plate capacitor, E is a constant, so the energy density is the same everywhere.
In general, the energy density will vary with space.

Please show your working for a cylindrical capacitor.

 

[onurbeyaz]

Thanks for the reply, I found this page from the link you gave me;
http://www.ece.umd.edu/class/enee380-1.F2004/lectures/lecture11.htm

In this page 3-11.1 seems exactly what I needed but I couldn't understand what (∇.D) is. My guess is D can be charge but it doesn't make sense. Could you please tell this with a simpler math

 

[Simon Bridge]

I couldn't understand what (∇.D) is. My guess is D can be charge but it doesn't make sense. Could you please tell this with a simpler math

I'm afraid it does not get simpler than that. You need to learn more maths.
I can try to put it in terms of things you may already know...

$\vec\nabla\cdot\vec D$ is the divergence of the electric displacement field.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/maxeq.html
... it is equal to the free charge density. So we write: $$\vec\nabla\cdot \vec D = \rho_{free}$$
The displacement field is related to the electric field by $\vec D = \epsilon\vec E$ where $\small\epsilon$ is the electric permitivity.

So for a charge q at the origin, $$\vec D = \frac{q}{4\pi}\frac{\vec r}{r^3}$$
$\vec \nabla$ is the divergence operator.
In cartesian coordinates (using i-j-k notation) it is: $\vec\nabla = \hat \imath \frac{\partial}{\partial x} + \hat \jmath \frac{\partial}{\partial y} + \hat k \frac{\partial}{\partial z}$
... the divergence is kind-of like the gradient of a function but for vector fields.