Question

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?

Answer 1

Answer:

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

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

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 ∫  $\frac{dq}{r^2}$

            cos θ = x / r

             

using the Pythagorean theorem

            r = $\sqrt{x^2 + y^2}$

we substitute

            Eₓ = k ∫ $\frac{dq}{x^2+y^2} \ \frac{x}{\sqrt{ x^2+y^2} }$

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

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

the ring's electric field

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

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 = $\frac{dq}{dA}$

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 = $\frac{1}{2\pi r} \ \frac{dq}{dr}$

             dq = 2π σ r dr

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

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

             

if we use a change of variable

               dv = 2rdr

               v = r²

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

we integrate

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

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

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

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

 

             σ = Q/πR²

substitute

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

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

             

The two electric fields are

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

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

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