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?

* E_ring =

*E_ disk= 2kQ  

Explanation:

Let's start by finding the electric field of the charged ring

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 ∫  

            cos θ = x / r

using the Pythagorean theorem

            r =

we substitute

            Eₓ = k ∫

            Eₓ =     ∫ dq

             Eₓ = k \frac{x}{(c^2+y^2)^{3/2} }  Q

the ring's electric field

             E_ring =

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 =

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 =

             dq = 2π σ r dr

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

             Eₓ =

if we use a change of variable

               dv = 2rdr

               v = r²

              Eₓ =  

we integrate

              Eₓ = k x π σ  

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

              Eₓ = k π σ x  (-2) [ ]

              Eₓ = 2π k  σ (  )

             σ = Q/πR²

substitute

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

             E_ disk= 2kQ  

The two electric fields are

* E_ring =

*E_ disk= 2kQ  

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

1 a

2 b

3 b

4 d

5 d

6 d

explanation: