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![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Apr 2008 00:20:12 IST
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Q1 Three identical charged small spheres each of mass m are suspended from a common point by insulated light string each of length l.The spheres are always on vertices of equilateral triangle of side length x<<<l .Calculate the rate dq/dt with which charge on each sphere increase if length of the sides of triangle increases according to law dx/dt = a/ x
Q2 Two mutually perpendicular long straight conductors carrying uniformly distributed charges of linear charge densities `1 and 2 are positioned at distance a from each other.How does the interaction between the rods depend on a?
PLZZ help!!!!!!
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"You need a perfect strategy to clear JEE more than knowledge " |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Apr 2008 10:05:25 IST
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Q2 independant of a
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Apr 2008 10:16:45 IST
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for wire 2
let x be distance of a piece of thickness dx from the point where they are perpendicular.
Then E1(x) = (lambda1)/2pi*epsilon*root(a^2+x^2)
then force on dx = (lambda1*lambda2*dx)/2pi*epsilon*root(a^2+x^2)
now horizontal component cancel and vertical remain
so dF(x) = (lambda1*lambda2*dx*a)/2pi*epsilon*(a^2+x^2)
evaluate integral from 0 to infinity it comes independant of a
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Apr 2008 10:32:27 IST
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q1 . Draw the FBD of any of the charged ball
Assume that the string connected to it makes an angle with the vertical .
Then we have T cos = mg
T sin = q^2/x^2 sqrt ( 3 ) ( resultant of two forces )
so , tan = q^ 2sqrt ( 3 ) / x^2 mg = x / sqrt(3 ) l ( from geqmetry with approximation )
solving for q , we get q = k x^(3/2)
or dq/dt = 3/2k sqrt ( x )dx/dt = c ( since dx/dt = a/ sqrt(x ) )
so , q = ct , c being a constant ( determine it by putting the value of k )
The calculation is done in CGS units .
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Apr 2008 23:57:08 IST
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feymann cant understand your solution
Plus your answer is also wrong
Plzzzzz try again and explain clearly.........
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Give the answer .
The method is clear enough .
But put clearly the value of dx/dt . Can't read properly .
I have taken that to be a/sqrt(x ) . In that case , the answer is pretty correct !!
The answer ( more precisely the value of c) would be different ( a factor of 4pi epsilon naught would be present ) if the calculation is done in SI method .But in that case also q must be proportioal to time .
Also the qn is incomplete !! It has not mentioned what initial charge was present in the balls . I have taken that to be zero .
Are u having problem with the value of c ? Let me know !
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Apr 2008 09:29:02 IST
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waiting for an answer !!!
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Apr 2008 12:54:03 IST
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Apr 2008 13:56:45 IST
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exactly ! ( I have not put the value of constant c , left as a problem (: )
See , u also get dq/dt = constant , just like me .
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 1 May 2008 00:11:34 IST
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OK OK feynmann gr888888 job
thanxxxxxxxx
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