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How much work is done in separating an electron (see Exercise 7 ) and an oxygen nucleus, which has a positive charge of $1.3 \times 10^{-18} \mathrm{C},$ from a distance of $2.0 \mu \mathrm{m}$ to a distance of $1.0 \mathrm{m} ?$

Calculus 2 / BC

Chapter 26

Applications of Integration

Section 6

Other Applications

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this problem we are asked how much work is done in separating an electron and an oxygen nucleus, which has a positive charge of 1.3 times 10 to the power of negative 18 columns from a distance of 2.0 micrometers to a distance of one m. The first thing that will note is that to micro meters is equal to two times 10 to the power of negative six m. Now we have that the force between two charged particles is given by K times Q one times Q two divided by X squared where X is the distance between them? Where I'll note that the positive charge of our oxygen nucleus is what I'll call Q one Q two will be the charge of an electron, which we can look at. The previous problem for So Q two is equal to negative. Where is it? Would be negative 1.6 times 10 to the power of -19. Cool owns. So the overall work done here is going to be equal to que que one q two times the integral from 2.0 times 10 to the power of negative six m up to one m Of X. to the power of -2 DX which would then be equal to K times Or excuse me, it be negative K times Q one times Q two times X to the power of negative one, evaluated from two times 10 to the power of negative six up to one m, which will give us then. So we'll find that the final result here will be approximately negative 9.36 times 10 to the power of negative 22 jewels.

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