CBSE Class 12-science Answered

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