Answer:
 E = 0    r <R₁
Explanation:
If we use Gauss's law
       Ф = ∫ E. dA =  / ε₀
 / ε₀
in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.
Consequently by Gauss's law the electric field is ZERO
            E = 0    r <R₁
 
        
             
        
        
        
It will be 
E = mgh.
where h and g are constant thus 
m can be written as 4/3πr^3*density 
E = 4/3πr^3* density 
E? = 4/3π(2R)^3* density 
 = 4/3π8r^3 
thus the e will be 4/3π8r^3* density/4/3πr^3*density nd thus you get 8E ..
        
             
        
        
        
Risks are only in a persons 
        
             
        
        
        
First we can say that since there is no external force on this system so momentum is always conserved.




now by the condition of elastic collision
![v_{2f} - v_{1f} = 0.8 - 0[\tex]now add two equations[tex]3*v_{2f} = 1.6](https://tex.z-dn.net/?f=v_%7B2f%7D%20-%20v_%7B1f%7D%20%3D%200.8%20-%200%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3Enow%20add%20two%20equations%3C%2Fp%3E%3Cp%3E%5Btex%5D3%2Av_%7B2f%7D%20%3D%201.6)

also from above equation we have

So ball of mass 0.6 kg will rebound back with speed 0.267 m/s and ball of mass 1.2 kg will go forwards with speed 0.533 m/s.