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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Mar 2008 15:58:49 IST
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y do we represt as electric as [q ] [ o] f/q??
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13 Mar 2008 17:30:23 IST
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it is when a test charge q is represented as q and is taken to Q (that is positive charge....) and so that q does not affect or apply any force q->0 and hence electric field ....i.e force per unit charge is represented as f/q and correctly represneted as q0 f/q
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adversities cause some men to break other to break records............i m of the other type....... :-) |
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13 Mar 2008 18:08:54 IST
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it is becoz if q (test charge) has mag. > 0 then the source charge will be displaced i.e moved due to the electric field of the test charge..and the net field due to source charge will not be correct.....so we take the mag. of the test charge as tends to zero (which is not possible....we only assume it)....ok
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13 Mar 2008 18:20:58 IST
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So that the charge in consideration has negligible electric field of its own, as compared to the field in which it is being moved
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- Gaurav Ragtah (spideyunlimited)
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15 Mar 2008 08:54:32 IST
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Perfect answer by everyone. The idea is that test charge should not disturb the configration of the system.
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Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
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