# Homework Help: Please explain this step

1. Apr 7, 2008

### name

Last edited by a moderator: May 3, 2017
2. Apr 7, 2008

### HallsofIvy

It's a little hard to distinguish between $\nabla$ and V!!
What you are asking is how they went from
$$\int\int\int (\nabla\cdot E)V d\tau$$
to
$$\int\int\int E\cdot\nabla V d\tau+ \int\int V (E\cdot dA)$$
(some people prefer $E\cdot n dA$ rather than $E\cdot dA$ where "n" is the unit normal to the surface.) V here is the scalar potential and E is a vector function.

There are actually two steps in there. First they are using the "product" rule:
[tex]\nabla\cdot(VE)= V\nabla\cdot E+ (\nabla V)\cdot E[/itex]
where $\nabla\cdot E$ and $\nabla\cdot (VE)$ are the divergence (div) of the vectors and $\nabla V$ is grad V.

so
[tex]\int\int\int\nabla\cdot(VE)d\tau= \int\int\int V\nabla\cdot Ed\tau +\int\int\int(\nabla V)\cdot E d\tau[/itex]

Now use the divergence theorem to convert that first integral on the right to
[tex]\int\int VE\cdot dA[/itex]

But I'm not at all clear why the $\epsilon_0/2$ only multiplies the first integral!