Equation connecting potential and potential energy of a distribution

[Leo Liu]

Homework Statement

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Relevant Equations

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The equation below allows us to calculate the potential energy of a continuous distribution of electric charge.
$$U=\frac {\epsilon_0} 2 \iiint\limits_\text{Entire electric field}\vec E^2\,dV$$
In my textbook, the author states
$$U=\frac 1 {8\pi\epsilon_0}\iiint\limits_\text{Entire electric field}\rho\phi\, dV$$
which relates the potential energy of the distribution to its electric potential. I wonder how it is derived from the first equation.

Edit: It looks like the equation is actually a special case of the following formula for a discrete configuration of charges:
$$U=\frac 1{4\pi\epsilon_0}\frac 1 2\sum_{j=1}^Nq_j\sum_{k\neq j}\frac{q_k}{r_{jk}}$$

 

[ergospherical]

The trick is to write $\displaystyle{\int} E^2 dV = - \displaystyle{\int} \mathbf{E} \cdot \nabla \phi dV = -\displaystyle{\int}\nabla \cdot (\mathbf{E} \phi) dV + \displaystyle{\int} \phi \nabla \cdot \mathbf{E} dV$, where the integral is taken over all space. The first term vanishes by Gauss' theorem since $\mathbf{E}$ is zero at infinity.

 

[Leo Liu]

The trick is to write $\displaystyle{\int} E^2 dV = - \displaystyle{\int} \mathbf{E} \cdot \nabla \phi dV = -\displaystyle{\int}\nabla \cdot (\mathbf{E} \phi) dV + \displaystyle{\int} \phi \nabla \cdot \mathbf{E} dV$, where the integral is taken over all space. The first term vanishes by Gauss' theorem since $\mathbf{E}$ is zero at infinity.

Can you tell me what the indentity $$\vec E\cdot\nabla\phi=\nabla\cdot(\vec E\phi)+\phi\nabla\cdot\vec E$$ is called? I am not very familiar with the del operator. Ty!

 

[ergospherical]

In suffix notation (with the $x_i$ Cartesian coordinates),\begin{align*}
\nabla \cdot (\mathbf{E} \phi) = \sum_i \dfrac{\partial}{\partial x_i} \left( E_i \phi \right) &= \sum_i \phi \dfrac{\partial E_i}{\partial x_i} + \sum_i E_i \dfrac{\partial \phi}{\partial x_i} \\

&= \phi \nabla \cdot \mathbf{E} + \mathbf{E} \cdot \nabla \phi
\end{align*}

 

[Steve4Physics]

Can you tell me what the indentity $$\vec E\cdot\nabla\phi=\nabla\cdot(\vec E\phi)+\phi\nabla\cdot\vec E$$ is called?

May I add (since you asked for a name) that ∇•(Eφ) = E•∇φ + φ∇•Eis a ‘product rule’ (e.g. see https://en.wikipedia.org/wiki/Vector_calculus_identities#Product_rule_for_multiplication_by_a_scalar )

Rearranged it gives -E•∇φ = -∇•(Eφ) + φ∇•E which is what @ergospherical used in their elegant solution.

 

[Leo Liu]

May I add (since you asked for a name) that ∇•(Eφ) = E•∇φ + φ∇•Eis a ‘product rule’ (e.g. see https://en.wikipedia.org/wiki/Vector_calculus_identities#Product_rule_for_multiplication_by_a_scalar )

Rearranged it gives -E•∇φ = -∇•(Eφ) + φ∇•E which is what @ergospherical used in their elegant solution.

This page is quite useful as it has all the identities of vector calculus. Thanks.