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reddevil_2009 (933)

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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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sagarvaze (253)

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Q2 independant of a






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sagarvaze (253)

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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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feynmann (1959)

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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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reddevil_2009 (933)

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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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feynmann (1959)

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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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feynmann (1959)

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waiting for an answer !!!

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elastiboysai (2327)

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\mbox{The 3 charges will be 3 vertices of a tetrahedron }\\ \mbox{with the fourth vertex as the point of suspension}\\<br/>\mbox{Draw the perpendicular bisector of one side}\\<br/>\mbox{ from the point of suspension}\\<br/>\frac{F_(electr.)}{F_(gravit.)}=\frac{x/2}{\sqrt{{l^2}-\frac{x^2}{4}}}\\<br/>\mbox{approximate it as}\frac{x}{2l} (x<<l)\\<br/>\frac{2q^2}{4 \pi \epsilon_0 x^2 mg}=\frac{x}{2l}\\<br/>x=q^\frac{2}{3}(\frac{l}{\pi \epsilon_0 mg})^\frac{1}{3}\\<br/>\mbox{Now for some manipulation}\\<br/>\frac{dx}{dt}=\frac{dx}{dq} \cdot \frac{dq}{dt}\\<br/>\frac{dq}{dt}=\frac{a}{\frac{dx}{dq} \cdot \sqrt{x}}\\<br/>\frac{dx}{dq}=(\frac{l}{\pi \epsilon_0 mg})^\frac{1}{3}q^\frac{1}{3}\\<br/>\sqrt{x}=(\frac{l}{\pi \epsilon_0 mg})^\frac{1}{6}q^\frac{1}{3}\\<br/>\mbox{Substitute these and ull find}\\<br/>\frac{dq}{dt}=\frac{3a}{2}\sqrt{\frac{\pi \epsilon_0 mg}{l}}\\

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feynmann (1959)

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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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reddevil_2009 (933)

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OK OK feynmann gr888888 job


 


thanxxxxxxxx


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