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Two positive charges of magnitude q are placed at the ends of a side (side 1 )of square of side 2a. Twonegative cahrges of the same magnitude ar kept at the other comers. Starting from rest, if the charge Q moves from the middle of side (1)to thecentre of square , its kinetc energy at the centre of square is |
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Answer» `1/(4 PI in_(0)) (2qQ)/a (1 - 1/(sqrt(5))) ` AC = AD = a ` AE = AF = sqrt(a^(2)+ (2a)^(2))` `= sqrt(5) a ` `BC = BD= BE= BF = sqrt2 a` Potential at pointA is ` V_(A)( 1/(4 pi in_(0)) q/(AC)+ 1/(4 pi in_(0)) q/(AD) -1/(4 pi in_(0)) q/(AE) - 1/(4 pi in_(0)) q/(AF))` `= 1/(4 pi in_(0) q/a + 1/( 4 pi in_(0)) q/a - 1/(4 pi in_(0)) q/(sqrt(5a)) - 1/(4 pi in_(0)) q/(sqrt(5a)))` `V_(A) = 2 (1/(4 pi in_(0) q/a)) -2 1/(4 pi in_(0)) q/(sqrt(5a)) = 1/(4 pi in_(0)) (2q)/a ( 1- 1/sqrt(5))` Potenital at pointB is ` V_(B)( 1/(4 pi in_(0)) q/(BD)+ 1/(4 pi in_(0)) q/(BE) -1/(4 pi in_(0)) q/(AE) - 1/(4 pi in_(0)) q/(BF))` ` (1/(4 pi in_(0))q/sqrt(2a) + 1/(4 pi in_(0)) q/(sqrt2a)- 1/(4 pi in_(0)) q/(sqrt(2 a)) - 1/(4 pi in_(0)) q/(sqrt(2a))` `V_(B)= 2 1/( 4 pi in_(0) q/(sqrt(2)a))- 2 (1/(4 pi in_(0))q/(sqrt(2)a) = 0` ` thereforeV_(A)-V_(B)= 1/(4 pi in_(0) (2q)/a (1-1/sqrt5)` Kineticenergyat the centre of squareof charge Q = WORK done on CHARGEQ fromA to B is ` W = Q (V_(A)-B-V_(B)) = 1/( 4pi in_(0)) ( 2 qQ)/a ( 1- 1/sqrt(5))` 
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