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raju manga Grade: 12
        

consider a uniformly charged ring of radius R.Find the point on the axis where the elecrtic field is maximum.

6 years ago

Answers : (3)

vikas askiitian expert
510 Points
										

let the total charge on the ring is Q then


electric field at the axis of a charged ring is given by


      E=kQx/(R2 + x2)3/2                


   to find maximum electric field we can use the concept of maxima and minima....


 dE/dx = d/dx of {kQx/(R2 + x2)3/2 }


          =kQ{  1 .(R2 +x2)3/2 - 3/2 (R2 +x2)1/2. x }/(R2 +x2)3                 (by applying quotient rule)


now on putting dE/dx =0 ,we get


      x2 -3x/2+R2 =0


      x2 -3x/2 +9/16  -9/16  +R2  =0            


       (x-3/4)2 = 9/16-R2


        x={(9-16R2)1/2 +3 }/4   units


 therefore at a distance x from center of the ring electric field is maximum.......

6 years ago
Askiitians_Expert Yagyadutt
askIITians Faculty
74 Points
										

Hello raju....


 


This is very simple...Just it needs a few steps of derivation...I am solving it with derivation...from the begining Let me know if u get stick with any line ....


 


It is :


Electric field due to:


 Charged ring of charge Q and radius R.

1) The Center=0.
Derivation:

Linear charge density (l) on ring =

Consider the field at center due to any element =

But the field due to point diametrically opposite = in opposite direction.
Net field at center = 0 (By symmetry)


2)      On the axis =


On axis of ring at distance x.
Derivation:
         
                                  Fig (5)


As obvious from the diagram the field component along the line gets added due to opposite element.




 


 


Now ...We have to find at which  x the field is maximum...so calculate dE/dx ...


 


dE/dx = Z [ (R^2 + x^2)^3/2  - 3/2(R^2+x^2)^1/2 . 2(x).(x) ] / [R^2+x^2]^3   ; where Z = Q/4pi*epsilon


 


For maximum let dE/dx = 0


 


then ..(R^2x +x^2)^3/2 - 3x^2(R^2+x^2)^1/2 = 0


(R^2 +x^2)^1/2 [ (R^2 +x^2) - 3x^2 ] = 0


 


hence ...R^2 = 2x^2


 


or .... x = + R/root(2)   or - R/root(2)


 


Means If ring is kept symmetrically in x-y plane with center at origin then..field will be maximum at ...a distance ..of +R/root(2) above the +z-axis...and below the negative z axis...


 


Is it clear


 


With regards


Yagya


askiitians_expert

6 years ago
T Purushotham
37 Points
										

363_89476_Screenshot-1.jpg

3 years ago
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