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Please solve the question 

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Asked by anshuman.anshuman090 30th October 2019, 12:24 PM
Answered by Expert
Answer:
To get the velocity of charge at surface of charged sphere, we equate the change in potential energy of charge from point P
to a point at surface of spher to change in kinetic energy of charge from point P to the point on the surface of sphere.
 
change in potential energy  ΔU is given by
 
begin mathsize 14px style increment U space equals space fraction numerator negative q space Q over denominator 4 pi epsilon subscript o left parenthesis 2 R right parenthesis end fraction space minus fraction numerator negative q space Q over denominator 4 pi epsilon subscript o left parenthesis R right parenthesis end fraction space space equals space fraction numerator q space Q over denominator 4 pi epsilon subscript o left parenthesis 2 R right parenthesis end fraction end style  .............................(1)
hence velocity at a point on surface of charged sphere :-  ΔU = (1/2)mv2 or velociy  begin mathsize 14px style v space equals space square root of fraction numerator q space Q over denominator 4 space pi space epsilon subscript o space m space R end fraction end root end style 
---------------------------------------------------------------
 
Change in potential from a point on surface of charged sphere to centre of sphere is calculated as follows
 
Electric field Er = - [ qQ / (4πεo) ] ( r/R3)
potential difference begin mathsize 14px style increment U space equals space minus integral E subscript r d r space equals negative integral subscript R superscript 0 fraction numerator negative q Q over denominator 4 πε subscript straight o end fraction r over R cubed d r space equals space minus fraction numerator q Q over denominator 4 πε subscript straight o end fraction fraction numerator 1 over denominator 2 R end fraction end style
Hence if we equate change in potential energy to change in kinetic energy
 
begin mathsize 14px style negative fraction numerator q space Q over denominator 4 πε subscript straight o end fraction fraction numerator 1 over denominator 2 R end fraction space equals space 1 half space m space open square brackets v subscript f superscript 2 space minus space v subscript i superscript 2 close square brackets
v subscript f superscript 2 space equals space v subscript i superscript 2 space minus space fraction numerator q space Q over denominator 4 πε subscript straight o end fraction fraction numerator 1 over denominator m space 2 R end fraction space equals space fraction numerator q space Q over denominator 4 πε subscript straight o end fraction fraction numerator 1 over denominator m space R end fraction minus space fraction numerator q space Q over denominator 4 πε subscript straight o end fraction fraction numerator 1 over denominator m space 2 R end fraction space equals space space fraction numerator q space Q over denominator 4 πε subscript straight o end fraction fraction numerator 1 over denominator m space 2 R end fraction end style
 
Hence velocity of charge at centre ,
 
begin mathsize 14px style v space equals space square root of fraction numerator q Q over denominator 8 πε subscript straight o straight m space straight R end fraction end root end style

begin mathsize 14px style negative fraction numerator q space Q over denominator 4 πε subscript straight o end fraction fraction numerator 1 over denominator 2 R end fraction space equals space 1 half space m space open square brackets v subscript f superscript 2 space minus space v subscript i superscript 2 close square brackets
v subscript f superscript 2 space equals space v subscript i superscript 2 space minus space fraction numerator q space Q over denominator 4 πε subscript straight o end fraction fraction numerator 1 over denominator m space 2 R end fraction space equals space fraction numerator q space Q over denominator 4 πε subscript straight o end fraction fraction numerator 1 over denominator m space R end fraction minus space fraction numerator q space Q over denominator 4 πε subscript straight o end fraction fraction numerator 1 over denominator m space 2 R end fraction space equals space space fraction numerator q space Q over denominator 4 πε subscript straight o end fraction fraction numerator 1 over denominator m space 2 R end fraction end style
Answered by Expert 1st November 2019, 1:48 PM
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