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![]() | Lec 4 | MIT 8.02 Electricity and Magnetism, Spring 2002 then that this equals Q one, Q two, divided by four pi epsilon zero. And now I have downstairs here an R squared. And so I have the integral now dR divided by R squared, from capital R to infinity. And this integral is minus one over R. Which I have to evaluate between R and infinity. And when I do that that becomes plus one over capital R. Right, the integral of dR over R squared I'm sure you can all do that is minus one over R. I evaluate it between R and infinity and so you get plus one ... |

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