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 Share with your friends Share 2 Koka Sri Lakshmi Divya Sai answered this Dear Student GivenE=qxa2+x232We have to find at which x the field is maximum for thatwe have to calculate dEdx⇒dEdx=qa2+x232-32a2+x212.2.x.xa2+x2322For maximum let dEdx=0⇒a2+x232-32a2+x212.2.x.x=0⇒a2+x232-3x2a2+x212=0⇒a2+x232=3x2a2+x212⇒a2+x2=3x2⇒2x2=a2⇒x2=a22⇒x=±a2So,E is maximum for x=±a2 on rthe axis from the centre of the ring. Regards 1 View Full Answer Prashant Kumar Giri answered this x=+-a. 0 Suranyo Chattopadhyay answered this Actually the amswer booklet is showing the answer to be option 1 -2