Four particles each of mass mass charge q are held at the vertices of a square of side a. They are released at t=0 and move under mutual repulsive forces. Find the speed of any paticle when its distance from the centre of the square is double.

Asked by Archit | 31st May, 2015, 07:14: PM

Expert Answer:

$begin mathsize 12px style Four space particles space each space of space mass space mass space charge space straight q space are space held space at space the space vertices space of space straight a space square space of space side space straight a. space To space find space the space speed space of space any space paticle space we space need space to space figure space out space intial space potential space energy space and space final space potential space energy space when space the space distance space of space square space is space double. Total space potential space energy space is space given space as comma increment straight U equals straight U subscript straight i minus straight U subscript straight f straight U subscript straight i equals Kq squared open square brackets 4 over straight a plus fraction numerator 2 over denominator square root of 2 straight a end fraction close square brackets straight U subscript straight f equals Kq squared open square brackets fraction numerator 4 over denominator 2 straight a end fraction plus fraction numerator 2 over denominator square root of 2 left parenthesis 2 straight a right parenthesis end fraction close square brackets From space the space figure space we space have space four space side space of space square space and space two space side space across space digonal space hence comma increment straight U equals Kq squared open square brackets 4 over straight a plus fraction numerator 2 over denominator square root of 2 straight a end fraction close square brackets minus Kq squared open square brackets fraction numerator 4 over denominator 2 straight a end fraction plus fraction numerator 2 over denominator square root of 2 left parenthesis 2 straight a right parenthesis end fraction close square brackets space space space space space space space space equals Kq squared open square brackets 4 over straight a plus fraction numerator 2 over denominator square root of 2 straight a end fraction minus fraction numerator 4 over denominator 2 straight a end fraction minus fraction numerator 2 over denominator square root of 2 left parenthesis 2 straight a right parenthesis end fraction close square brackets increment straight U equals Kq squared open square brackets 2 over straight a plus fraction numerator 1 over denominator square root of 2 straight a end fraction close square brackets To space find space the space speed space of space any space particle space we space need space to space divide space potential space energy space by space 4. since space potential space enegy space loss space by space 4 space charges. By space the space conservation space of space energy space loss space in space potential space energy space is space equal space to space kinetic space energy. 1 fourth Kq squared open square brackets 2 over straight a plus fraction numerator 1 over denominator square root of 2 straight a end fraction close square brackets equals 1 half mv squared straight v squared equals Kq squared over straight m open square brackets 2 over straight a plus fraction numerator 1 over denominator square root of 2 straight a end fraction close square brackets The space velocity space of space any space particle space will space be comma straight v equals square root of Kq squared over straight m open square brackets 2 over straight a plus fraction numerator 1 over denominator square root of 2 straight a end fraction close square brackets end root end style$

Answered by Priyanka Kumbhar | 1st Jun, 2015, 03:00: PM

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