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

```							Let the total charge on the ring is Q thenelectric 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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10 months ago
```							let the total charge on the ring is Q thenelectric 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
```
9 months ago
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