Joint Entrance Examination/AIEEE 2009/Q084 work done by electrostatic field
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Original Question by 203.199.213.66
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Statement 1 : For a charged particle moving from point P to point Q, the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q. Statement 2 : The net work done by a conservative force on an object moving along a closed loop is zero.
Posted by 203.199.213.66 13:40, 22 August 2009 (UTC) |
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If you notice, the two statements that are given are textbook statements. Textbooks always use these statements to describe an electrostatic field and a conservative field. This essentially means that the student is familiar with the two statements (unlike JEE Assertion/Reason statements). So the question here is, does statement 2 explain statement 1? I'll point out one thing: - Statement 2 is about Conservative fields and loops; while statement 1 is about electrostatic fields and paths. This is what the question focuses on, according to me. How will work across a loop being zero imply that work done is same across any path? Is an electrostatic field always conservative? I know it's an insult to students to ask them the last question. It's THAT simple. But still, people will have doubts. After all, an induced electric field is NOT conservative! but wait a minute! Is an induced electric field electrostatic? This is how we are supposed to learn - ask yourself as many questions as you can... you want to clear up the basics. So, carry on with the answer and the discussion. |
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This will give you a general idea of how the discussions on this site will go. I enjoy it and now I'll have the added satisfaction that people will actually read it! Dude, register on the site, that'll make sure that when you post questions, your name will show up and NOT your IP address! As for the maths thing you asked... <math>\frac{a}{b}</math>
<math>\frac{ax^2 + bx + c}{9}</math>
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