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Vector / Integral question - probably simple, but can't see answer.

http://www.maths.ox.ac.uk/system/files/coursematerial/2012/2393/15/Sheet3_13.pdf

It's question 6 of the above, just the bit at the start of step c) where you prove

\nabla\phi(r)=\int \frac{k\nabla'\rho(r')}{|r-r'|}dV'

.

Basically, you have

\phi(r)

, and you can assume the identities proved in a) and b). Prime on a vector operator means treat r' as variable and r as constant.

What I did was;

\nabla\phi(r)=\int\nabla(\frac{k\rho(r')}{|r-r'|})dV'

=\int\nabla(\frac{1}{|r-r'|})k\rho(r')dV'

=-k\int\nabla'(\frac{1}{|r-r'|})\rho(r')dV'

Then using integration by parts or the product rule;

=-k\int\nabla'(\frac{\rho(r')}{|r-r'|})\rho(r')dV'+k\int(\frac{ \nabla'(\rho(r'))}{|r-r'|})dV'

However, this is the wrong answer, as the first term won't always be zero and the second term is the required integral. The class notes I copied off the board (without understanding at the same time) seem to do the same thing I have, but the answer is definitely the one asked for, as that's what it says in the textbook.

Can anyone tell me what I'm missing. I don't know where to start looking. Thanks :smile:

EDIT: I've solved the problem - there is a typo in the question. It should say rho vanishes outside and on the boundary, which causes the extra term to disappear, by the divergence theorem. I am now really happy. Feel free to delete the thread. :smile:

(edited 10 years ago)

Reply 1

jxsoph

"probably simple"