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

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.

d. platinum would have the greatest temperature increase because it has the lowest specific heat.