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```				   consider a uniformly charged ring of radius R.Find the point on the axis where the elecrtic field is maximum.
```

6 years ago

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```										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
```										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.Derivation: 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.Derivation:                                                                          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 maximum...so 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
Yagya
```
6 years ago
```
```
2 years ago

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