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

## 2 Answers

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

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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