Question

11.
The maximum electric field intensity on the axisof a uniformly charged ring of charge q and radius
R will be :

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Answer 1

The maximum electric field intensity is at a distance of R/root(2) from center and on axis

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Answer 2

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