The electric filed at a distance x from the center of a uniformly charged disc of radius R,along the axis passing through the center is given by E = sigma/2e(1-x/rt(x^2+R^2)) where sigma is surface charge density and e is permittivity of free space .Putting x=0 in this eqn gives filed at the center of the disc =sigma/2e ?? How is this possible ??? By symmetry considerations and considering the ring to be made up of concentric rings; for each such ring field at the center will be zero and thus the net field must be zero?? Can anyone explain why it is so ??

- anonymous

- schrodinger

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

help @electrokid

- anonymous

Gauss's Law is the answer :)
|dw:1366039832463:dw|

- anonymous

actually i know that but a task is given to me to think abt it without using gauss's law.......

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

then use the fact that since the outside of the disc is positively charged, due to induction the inside would be negatively charged.
Then take the intensity at the center due to an infinitesimal element of the disc and then integrate it throughout
the resulting vector sum = 0

- anonymous

so, you are indirectly asked to proove Gauss's Law

- anonymous

that is what the question is actually according to symmetry field at centre should be zero but actuallt it is not.... and the charge is uniformly distributed over the disk with surface charge density (say sigma) so the whole disk is positively charged....

- anonymous

the only time when the inside of a sphere has non-zero intensity is when it is a non-conductor or an insulator.

- anonymous

it is a disk...... and since charge is uniformly distributed all over the disk therefore obviously it is an insulator

- anonymous

please help buddy....

- anonymous

@.Sam.

- anonymous

ok.. look at the denominator of the equation..
rewrite the equation for E using the equation editor

- anonymous

\[\sigma/2\pi \epsilon(1-\chi/\sqrt{r ^{2}+\chi ^{2}}) \] where sigma is surface charge density, x is dist of from centre on the axis of disk..... now puttin x=0 we are not gettin zero but considerin symmetry it should be zero... why is it so????

- anonymous

@Vincent-Lyon.Fr

- anonymous

|dw:1366042440302:dw|

- anonymous

that symmetry argument is invalid!~

- anonymous

i am sorry but i have told before and even now i am telling you that since charge is uniformly distributed over the whole disk therefore it can not be a conductor and hence it is an insulator |dw:1366077097337:dw| and the charge is positive...

- anonymous

??????

- anonymous

|dw:1366077552708:dw| now the diametrically opposite points will cancel each other's field and the field at centre should be zero.... but keeping x=0 in the equation we don't get electric field zero... why is it so????
hope now u understand my question....

- Vincent-Lyon.Fr

The formula given is only valid for x>0, not for x=0.
This is because there is a discontinuity at x=0 when you cross the charged sheet.
The field vs x looks like this:
|dw:1366043565515:dw|

- anonymous

but how to explain it just by concept of physics... i.e. physically and not mathematically....

- Vincent-Lyon.Fr

Finished drawing here:
|dw:1366043626646:dw|

- anonymous

so value of field at centre is not zero... right????

- anonymous

???

- anonymous

?????? @Vincent-Lyon.Fr

- anonymous

mathematically its ok but how to explain it physically contradicting the symmetry criteria.????@Vincent-Lyon.Fr

- Vincent-Lyon.Fr

Value at centre is 0 because of symmetries.

- anonymous

are you sure????

- anonymous

but a question is given to me and it is asked that why field at centre is not zero...???

- Vincent-Lyon.Fr

They probably mean: at centre of disc, on the surface (not 'inside').
The field there is \(\sigma/2\epsilon_o\), because the disc is seen as a uniform plane when you get extremely close.

- Vincent-Lyon.Fr

- anonymous

but we are considering the disk to be a very thin disk and, using this assumption we reach to the expression used above(we are assuming that charge is only at one face of the disk)......

- Vincent-Lyon.Fr

Remember a 2D charge distribution is only a model.
A real distribution is always 3D, so there is no discontinuity in reality.
E-field goes smoothly from \(-\sigma/2\epsilon_o\) on one side to \(+\sigma/2\epsilon_o\) on the other side.

- anonymous

i know that......
but while deriving the equation we consider the charge to be distributed on the surface...
and if 2D arrangement would have been possible we should get field to be zero at the centre... but according to the equation we don't get field zero...(*THE QUESTION IS VALID BECAUSE WE HAVE CONSIDERED THE DISK TO BE A 2D OBJECT WHILE DERIVING THE EXPRESSION OF FIELD DUE TO A DISK....)

- anonymous

Hello My Dear
First things first, if the disk is uniform, there will be charge at its center too, so there's no way to get the field void at its center.
If there is a small hole, the first ring, oh yes we can say that the electric field is zero at its center.
Now, this equation was obtained summed up several rings to form a uniform disk, there is no hole in its center so there is no way to get the electric field at x = 0 by it.
In fact, you can estiamr when the electric field are near the surface of the disk x << r.
Will obtain,\[\sigma/2\pi \epsilon _{0} \]
Hope i helped you

- Vincent-Lyon.Fr

Actually : \(\sigma/2 \epsilon_o\) without the \(\pi\) in the denominator.

Looking for something else?

Not the answer you are looking for? Search for more explanations.