# Green's First Identity involving Electric Potential

1. Jun 23, 2014

### Parmenides

I am attempting to work through a paper that involves some slightly unfamiliar vector calculus, as well as many omitted steps. It begins with the potential energy due to an electric field, familiarly expressed as:
$$U_{el} = \frac{\epsilon_r\epsilon_0}{2} \iiint_VE^2dV = \frac{\epsilon_r\epsilon_0}{2} \iiint_V(\nabla{\Phi})^2dV$$
I am not quite sure as to the expression for a squared gradient, but I worked from using $\nabla(\Phi\nabla\Phi) = (\nabla\Phi)^2 + \Phi\nabla^2\Phi$ so that $(\nabla\Phi)^2 = \nabla(\Phi\nabla\Phi) - \Phi\nabla^2\Phi$. Now, I can rewrite $U_{el}$ as:
$$U_{el} = \frac{\epsilon_r\epsilon_0}{2}\Big[\iiint_V\nabla(\Phi\nabla\Phi)dV - \iiint_V\Phi\nabla^2\Phi{dV}\Big]$$
I recognize the presence of Poisson's equation in the second term, but as for the first? This is where I am stuck. I am supposed to use these expressions together with the Divergence theorem so that I can invoke Green's first identity:
$$\int_{\Omega}(\psi\nabla^2\phi + \nabla\phi\cdot\nabla\psi)dV = \oint_{\Omega}\psi(\nabla\phi\cdot{\bf{n}})dS$$
For two scalar functions $\psi$ and $\phi$. Ultimately, I need to arrive at the expression:
$$U_{el} = \frac{1}{2}\oint_S\sigma\phi_0dS + \frac{1}{2}\iiint_V\rho\Phi{dV}$$
Using Poisson's equation $\nabla^2\Phi = -\frac{\rho}{\epsilon_0}$ and $\sigma = -{\epsilon_r\epsilon_0}\frac{\partial{\Phi}}{\partial{n}}$ and using the fact that $\phi_0$ is the potential at the surface.

But I do not have any dot products associated with what I did above and so I do not see how to proceed; the only way I see how this could work is understanding that there is really only one scalar function here instead of two, namely $\Phi$, but I don't see this allowing me to make $\nabla(\Phi\nabla\Phi) = \nabla\Phi\cdot\nabla\Phi$. Am I forgetting something simple?

2. Jun 23, 2014

### Matterwave

You...you are right there to solving it. You already showed that $\nabla(\Phi\nabla\Phi)=(\nabla\Phi)^2+\Phi\nabla^2\Phi$

So I think the only point you're stuck on should be: $(\nabla\Phi)^2=\nabla\Phi\cdot\nabla\Phi$

Can you do it from here?

Maybe a slightly bigger hint. I can rewrite Green's identity, if $\phi=\psi$ as such:

$$\int_\Omega (\psi\nabla^2\phi+\nabla\phi\cdot\nabla\psi)dV=\int_\Omega (\phi\nabla^2\phi+\nabla\phi\cdot\nabla\phi)dV=\int_\Omega \nabla(\phi\nabla\phi) dV=\oint_{\partial\Omega} \phi(\nabla\phi\cdot{\bf{n}})dS$$

Last edited: Jun 23, 2014