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the maximum electric field intensity on the axis of uniformly charged ring of charge q and radious r will be

the maximum electric field intensity on the axis of uniformly charged ring of charge q and radious r will be 

Grade:12

2 Answers

Arun
25763 Points
one year ago
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
 
Khimraj
3007 Points
one year ago

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

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