Two fixed point charges +4e and +2e are separated by a distance of 3a. Where should a 3rd charge be placed for the system to be in equilibrium ?

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Solution

$x i s t h e d i s \tan c e f r o m s m a l l c h a r g e w h e r e E i s z e r o o r e q u i l i b r i u m, s o a t t h i s p o i n t e l e c t r i c f i e l d d u e t o b o t h t h e c h a r g e s i s i n o p p o s i t e d i r e c t i o n 0 . \frac{4 e}{(3 a - x)^{2}} = \frac{2 e}{x^{2}} (3 a - x)^{2} = 2 x^{2} 3 a - x = 1.414 x 3 a = 2.414 x x = \frac{70}{2.58} = 1.2 a$

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Two fixed point charges +4e and +2e are separated by a distance of 3a. Where should a 3rd charge be placed for the system to be in equilibrium ?

x is the distance from small charge where E is zero or equilibrium,so at this point electric field due to both the charges is in opposite direction0.4e(3a-x)2 = 2ex2(3a-x)2=2x23a-x= 1.414x3a= 2.414xx = 702.58 =1.2a