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- Thread starter rojasharma
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- #2

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Work is defined as F*ds. Look at the equations you used and the units they give.

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- #5

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Don;t see that problem in cramster.com:(, thanks for the web though.

- #6

Doc Al

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That sounds right. You need to find the change in electric potential energy as the charges are brought together.what i did was..used Ee1=kq1q2/r1 (r1 as 0.4m), Ee2=kq1q2/r2(r2 as 100m), then found delta Ee= Ee2=Ee1, Delta Ee= W??

- #7

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q1=3.2x10^-3C

q2=1x10^-6C

r1=0.4m

r2=100m

Solution

E1= 72J (E1= (kq1q2)/r1)

E2= 0.288J (E2=kq1q2/r2)

delta E= E2-E1

delta E= 0.288J - 180J

delta E=-179.712J

Delta E= work done...(what about the negative value??)

- #8

Doc Al

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Looks OK. (You seemed to have switched r1 & r2: the charge movesSo,

q1=3.2x10^-3C

q2=1x10^-6C

r1=0.4m

r2=100m

Solution

E1= 72J (E1= (kq1q2)/r1)

E2= 0.288J (E2=kq1q2/r2)

Where did you get 180J from? The negative sign is due to you mixing up initial and final positions.delta E= E2-E1

delta E= 0.288J -180J

delta E=-179.712J

Delta E= work done...(what about the negative value??)

- #9

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Oh, sorry the E1 is 72J.....so its delta E= E1- E2???/ how did i mix up?

- #10

Doc Al

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Initial position: r1 = 100 m.

Final position: r2 = 0.4 m.

Final position: r2 = 0.4 m.

- #11

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right ...silly me...thanks a lot

- #12

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That sounds right. You need to find the change in electric potential energy as the charges are brought together.

hmm. I am thinking of writing the force between the two charges and integrating it over the distance.

- #13

Doc Al

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Which is how one wouldI am thinking of writing the force between the two charges and integrating it over the distance.

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