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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.

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.
                                  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 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



6 years ago
T Purushotham
37 Points


3 years ago
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