Electric field Intensity due to a Uniformly Charged Ring - JEE PDF Download

Electric field Intensity due to a uniformly charged ring at its Centre

Here by symmetry we can say that electric field intensity at Centre due to every small segment on ring is cancelled by the electric field intensity at Centre due to the segment exactly opposite to it. The electric field intensity at Centre due to segment AB is cancelled by that due to segment CD. Thus net electric field intensity at the Centre of a uniformly charged ring is E₀ = 0

Electric field Intensity due to a uniformly charged ring at a point on it’s axis

Determining the electric field intensity at a specific point, denoted as P, on the axis of a uniformly charged ring is our next focus. In this scenario, we examine the electric field at point P, which is located at a distance x from the center "O" of the ring. To facilitate our analysis, we direct our attention to two diametrically opposite small sections of equal length dl (highlighted in blue) on the ring, as depicted in the figure below.

The charge on these elemental sections is given by
Where Q = total charge on the ring
Due to the element dl, electric field strength dE at point P can be given by

The component of this field intensity dE sinθ which is normal to the axis of ring will be cancelled out due to the ring’s diametrically opposite section’s field component dE’ sinθ.
The component of electric field intensity along the axis of ring dEcosθ due to all the sections will be added up. Hence total electric field intensity at point P due to the ring is given by :-

In vector form,