JEE Class main Answered

Asked by SanskarAgarwal86 | 30 May, 2019, 09:00: PM

Expert Answer

For system to be in equilibrium the net force on charges should be zero

Let us take the distance beween Q1 and Q2 be and q is placed at adistance x from the charge Q1 and therefore the distance between q and Q2 would be d-x.

N e t space F o r c e space o n space c h a r g e space Q subscript 1 fraction numerator K Q subscript 1 q over denominator x squared end fraction plus fraction numerator K Q subscript 1 Q subscript 2 space end subscript over denominator d squared end fraction equals 0 fraction numerator K Q subscript 1 q over denominator x squared end fraction equals negative fraction numerator K Q subscript 1 Q subscript 2 space end subscript over denominator d squared end fraction minus negative negative negative negative negative negative negative negative 1 S i m i l a r l y space f o r space Q subscript 2 comma space end subscript t h e space n e t space f o r c e fraction numerator K Q subscript 2 q over denominator open parentheses d minus x close parentheses squared end fraction equals negative fraction numerator K Q subscript 1 Q subscript 2 space end subscript over denominator d squared end fraction minus negative negative negative negative negative negative negative negative 2 O n space c o m p a r i n g space e q u a t i o n space 1 space a n d space 2 fraction numerator K Q subscript 1 q over denominator x squared end fraction equals fraction numerator K Q subscript 2 q over denominator open parentheses d minus x close parentheses squared end fraction Q subscript 1 open parentheses d minus x close parentheses squared equals Q subscript 2 x squared O n space s i m p l i f y i n g space t h e space a b o v e space e q u a t i o n space w e space g e t space q u a d r a t i c space e q u a t i o n space i n space t e r m s space o f space x x squared open parentheses Q subscript 1 minus Q subscript 2 close parentheses minus 2 d x Q subscript 1 plus Q subscript 1 d squared equals 0 x equals fraction numerator 2 d Q subscript 1 plus-or-minus square root of 4 d squared Q subscript 1 squared minus 4 open parentheses Q subscript 1 minus Q subscript 2 close parentheses Q subscript 1 d squared end root over denominator 2 open parentheses Q subscript 1 minus Q subscript 2 close parentheses end fraction O n space s o l v i n g space x equals fraction numerator d square root of Q subscript 1 end root over denominator square root of Q subscript 1 end root plus square root of Q subscript 2 end root end fraction F r o m space t h e space f i r s t space e q u a t i o n space w e space g e t q equals negative Q subscript 2 x squared divided by d squared O n space s u b s t i t u t i n g space t h e space v a l u e space o f space x space w e space g e t space q equals fraction numerator negative Q subscript 1 Q subscript 2 over denominator open parentheses square root of Q subscript 1 end root plus square root of Q subscript 2 end root close parentheses squared end fraction T h e r e f o r e space q space s h o u l d space b e space n e g a t i v e space i n space p o l a r i t y space a n d space m a g n i t u d e space i s fraction numerator Q subscript 1 Q subscript 2 over denominator open parentheses square root of Q subscript 1 end root plus square root of Q subscript 2 end root close parentheses squared end fraction

Answered by Utkarsh Lokhande | 31 May, 2019, 09:50: AM

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