I f   E = q x ( a 2 + x 2 ) 3 2   r e p r e s e n t s   t h e   e l e c t r i c   f i e l d   o n   t h e   a x i s   o f   t h e   u n i f o r m l y   c h a r g e d   r i n g   o f   r a d i u s   ' a ' ,   t h e n   v a l u e   o f   E   i s   m a x i m u m   f o r   w h a t   v a l u e   o f   X   o n   t h e   a x i s   f r o m   t h e   c e n t r e   o f   t h e   r i n g ? ( 1 )   x = ± a 2 ( 2 )   x = ± a ( 3 )   x = ± a 2 ( 4 )   x = ± a 2 2

Dear Student

GivenE=qxa2+x232We have to find at which x the field is maximum for thatwe have to calculate dEdxdEdx=qa2+x232-32a2+x212.2.x.xa2+x2322For maximum let  dEdx=0a2+x232-32a2+x212.2.x.x=0a2+x232-3x2a2+x212=0a2+x232=3x2a2+x212a2+x2=3x22x2=a2x2=a22x=±a2So,E is maximum for x=±a2 on rthe axis from the centre of the ring.
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x=+-a.
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Actually the amswer booklet is showing the answer to be option 1
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