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In Griffiths' E&M, there is an equation that describes energy of a charge distribution as-

$W = \frac{\epsilon_0}{2}\int(\nabla.\textbf{E})V d\tau$

The author then performs integration by parts to get-

$W = \frac{\epsilon_0}{2}[-\int \textbf{E}.(\nabla V)d\tau + \oint V\textbf{E}.d\textbf{a} ]$

I understand that the right side of the equation comes from using the Divergence theorem, but I am unable to figure out how exactly the left side appears.

Paddy
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1 Answers1

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If $a$ is a vector field and $f$ a scalar function, then $$\mathrm{div}(f a)=f\mathrm{div}(a)+\nabla f \cdot a \, .$$ The previous one is a pointwise vector calculus identity. Then, integrate both sides and apply the divergence theorem to the left-hand side.

Kosh
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  • That works! But you aren't using integration-by-parts though, right? – Paddy Oct 22 '20 at 12:04
  • Well I am "proving" integration by parts formula. It is the name given to the integral formula that results by using divergence theorem in the way I wrote. – Kosh Oct 22 '20 at 12:54
  • Even in 1D you prove integration by parts as I wrote. It is nothing but the derivative of a product. Try to specialize what I wrote the the 1d case. – Kosh Oct 22 '20 at 13:40