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

509 Points
13 years 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.......

13 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

T Purushotham
37 Points
10 years ago

4 years ago
Dear student,

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 – x . 3/2 (R2 +x2)1/2. 2x }/(R2 +x2)3                 (by applying quotient rule)
now on putting dE/dx =0 ,we get
3x2 – (R2 + x2) = 0
or, 2x2 – R2 = 0
or, x = + R/√2 or x = – R/√2

Thanks and regards,
Kushagra