# Electric field inside a uniformly charged ring

It is known that electric field inside a uniformly positively charged ring (for points in the plane of the ring) is directed towards centre. But when i tried to prove this using the following steps, it turns out to be 0. Let us consider a point P in plane of a uniformly charged ring with centre at 0. Lets draw a diameter AB for the ring which contains the point P. Now, if we rotate the ring along this diameter AB, the locus followed by ring would be a spherical shell. If we assume that electric field at P is towards the ring's centre, rings obtained by rotating the ring along AB would also have the field towards centre O , as they all have a common centre O and for each ring, point P is equivalent location. By superimposition, sum of field due to all such rings (obtained by roation of the original ring) at P would be equal to field at P inside a uniformly charged shell,which is zero. Thus field by individual rings must be zero. But its not. Please tell me where i am mistaken.

• Why do you say it's not? It is, as you have elegantly proven. – Bzazz May 28 '17 at 7:59
• But it is already proven by other methods to be non zero. But this methods ends up in the field being zero. So whats the mistake? – user157353 May 28 '17 at 8:05
• Potentially you're not creating a uniformly charged shell when you rotate. But I don't know. – Bzazz May 28 '17 at 8:22

As Bzazz points out, the shell you are constructing has not its charge uniformly distributed. In fact the charge density follows the law $D_q(\vec r)=\dfrac{\lambda}{2\pi}\dfrac{1}{\sin\alpha}$, with $\vec r$ the position of a point in the shell, $\lambda$ is the density of charge per unit of length of the original ring and $\alpha$ is the angle formed by $\vec r$ and the axis of rotation you've chosen. Even not considering other issues, clearly the construction doesn't preserve the symmetry of the original problem, then it's not a surprise it doesn't give the expected result either.