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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 Apr 2008 20:53:20 IST
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Hey guys.....had a doubt........
Q. When we place three charges along a line....all three of same sign...then the charge in middle...will be in.... " stable/unstable " equlibrium ??
Book says..stable...
and similarly....
Q. When we place three charges along a line....extreme charges of same sign and middle one of opposite sign...then the charge in middle...will be in.... "stable/unstable " equlibrium ??
Book says...unstable..
BUT I HAVE A CONCEPTUAL REASON...to think the opposite...
pls post ur views...with proper comments....thanks guys in advance.......
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15 Apr 2008 21:07:51 IST
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neone out there???
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There's Light at the end of every Tunnel, so KEEP MOVING....
Best of luck to all my mates....
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15 Apr 2008 21:13:29 IST
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Q1
stable because say if you push it towards one charge that charge will exeret a larger force and so it will come back to central position where two charges ballance out.
You displace the charge anywhere on the line and it will come to mean position and stop.
Q2
unstable because if you give it a slight push to either side then one charge will exert a larger attractive force so it will pull the middle charge towards itself till they both join together.
Anyways what was your reasoning to think otherwise
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15 Apr 2008 21:23:04 IST
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ok...sagar,,,,agreed.......NOW HERE's my logic...
What if the middle charge is displaced in a direction perpendicular to the axis of the 3 charges....then????....
rest I leave upto u to think and reply....
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............................................................................................................................................................................................
There's Light at the end of every Tunnel, so KEEP MOVING....
Best of luck to all my mates....
............................................................................................................................................................................................
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15 Apr 2008 21:33:43 IST
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well actually its motion direction decides its stability and all.....now incase it is displaced along x axis....thenn its under stable equilib..and is in neutral equilib when displaced along y axis.,....hope its clear......
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15 Apr 2008 21:36:31 IST
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question states along a line but if it is displaced perpendicular then your answers are correct.
it seems that if you consider U(x) then that point comes as minima in 1st and maxima in 2nd but in U(y) graphs it is maxima in 1st and minima in 2nd
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15 Apr 2008 21:38:42 IST
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well what I feel....is....if it is displaced along y-axis....then...
it shud not be in neutral equlibrium but rather....it shud continue to move away from that position....
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............................................................................................................................................................................................
There's Light at the end of every Tunnel, so KEEP MOVING....
Best of luck to all my mates....
............................................................................................................................................................................................
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15 Apr 2008 21:40:05 IST
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thats wat is called neutral equilib......it will keep on moving...
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The only place where you will find success before hard work is in the dictionary
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15 Apr 2008 21:41:34 IST
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No buddy....had it been in neutral equilibrium....then it wud have remained at the same position where it was displaced...but its not so...
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............................................................................................................................................................................................
There's Light at the end of every Tunnel, so KEEP MOVING....
Best of luck to all my mates....
............................................................................................................................................................................................
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15 Apr 2008 21:43:02 IST
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edited...sorrry.....along y axis ..it will be in unstable equlib.....
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15 Apr 2008 21:46:50 IST
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just see this pic.......net force on it will be along y - axis and not zero.......
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............................................................................................................................................................................................
There's Light at the end of every Tunnel, so KEEP MOVING....
Best of luck to all my mates....
............................................................................................................................................................................................
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15 Apr 2008 21:48:13 IST
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chk the edited post.....i didnt chk...i just posted somethin......anyway...i corrected it....it will be in unstable equilib along y axis....
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