Assume a uniformly charged ring of radius R and charge Q produces an electric field Ering at a point P on its axis, at distance x away from the center of the ring. Now the charge Q is spread uniformly over the circular area the ring encloses, forming a flat disk of charge with the same radius. How does the field Edisk produced by the disk at P compare to the field produced by the ring at the same point?Assume a uniformly charged ring of radius R and charge Q produces an electric field Ering at a point P on its axis, at distance x away from the center of the ring. Now the charge Q is spread uniformly over the circular area the ring encloses, forming a flat disk of charge with the same radius. How does the field Edisk produced by the disk at P compare to the field produced by the ring at the same point?

in the attachment we can see a diagram of the system. Due to circular symmetry, the electric field perpendicular to the axis is canceled and only the electric field remains parallel to the axis.

Eₓ = E cos θ (1)

E = k ∫ [tex]\frac{dq}{r^2}[/tex]

cos θ = x / r

using the Pythagorean theorem

r = [tex]\sqrt{x^2 + y^2}[/tex]

we substitute

Eₓ = k ∫ [tex]\frac{dq}{x^2+y^2} \ \frac{x}{\sqrt{ x^2+y^2} }[/tex]

Eₓ = [tex]k \frac{x}{(c^2+y^2)^{3/2} }[/tex] ∫ dq

Eₓ = k \frac{x}{(c^2+y^2)^{3/2} } Q

the ring's electric field

E_ring = [tex]k \ \frac{x}{(x^2+ y^2)^{3/2} } \ Q[/tex]

Now let's find the electric field of the disk

The charge is distributed over the entire disk, so let's use the concept of charge density

σ = [tex]\frac{dq}{dA}[/tex]

Let's approximate the disk as a group of rings, the width of each ring is dr, the area is

dA = 2πr dr

we substitute

σ = [tex]\frac{1}{2\pi r} \ \frac{dq}{dr}[/tex]

dq = 2π σ r dr

we substitute in equation 1, where the electrioc field is of each ring

Eₓ = [tex]k \int\limits^R_0 \ { \frac{x}{(x^2+r^2)^{3/2} } \ 2\pi \sigma \ r } \, dr[/tex]

if we use a change of variable

dv = 2rdr

v = r²

Eₓ = [tex]k x \pi \sigma \int\limits^a_b { \frac{1}{(x^2+v)^{3/2} } } \, dv[/tex]

we integrate

Eₓ = k x π σ [tex][ \frac{ (x^2 + r^2)^{-1/2} }{-1/2} ][/tex]

we value in the limits from r = 0 to r = R

Eₓ = k π σ x (-2) [ [tex]\frac{1}{ \sqrt{x^2+R^2} } - \frac{1}{x}[/tex]]

Eₓ = 2π k σ ([tex]1 - \frac{x}{(x^2 + R^2 ) ^{1/2} }[/tex] )

σ = Q/πR²

substitute

Eₓ = 2 k Q/R² (1 - \frac{x}{(x^2 + R^2 ) ^{1/2} } )

E_ disk= 2kQ [tex]\frac{1}{R^2} \ (1 - \frac{x}{(x^2+ R^2)^{1/2} } )[/tex]

The two electric fields are

* E_ring = [tex]k \ \frac{x}{(x^2+ y^2)^{3/2} } \ Q[/tex]

*E_ disk= 2kQ [tex]\frac{1}{R^2} \ (1 - \frac{x}{(x^2+ R^2)^{1/2} } )[/tex]

we can see that the functional relationship of the two fields is different