A thin, uniformly charged spherical shell has a potential of 832 V on its surface. Outside the sphere, at a radial distance of 21.0 cm from this surface, the potential is 499 V. Calculate the radius of the sphere. Submit Answer Tries 0/32 Determine the total charge on the sphere. Submit Answer Tries 0/100 What is the electric potential inside the sphere at a radius of 4.0 cm? Submit Answer Tries 0/32 Calculate the magnitude of the electric field at the surface of the sphere. Submit Answer Tries 0/32 If an electron starts from rest at a distance of 21.0 cm from the surface of the sphere, calculate the electron's speed when it reaches the sphere's surface.

a

The radius is              

b

The total  charge is  

c

The electric potential inside sphere is the same with that outside the sphere which implies that the electric potential is 832 V

d

The magnitude of the electric field is  

e

The velocity is  

Explanation:

From the question we are told that

     The potential is

       The radial distance from the sphere is

       The potential at the radial distance is

The potential at the surface of the sphere is mathematically represented as

Where kQ is  a constant what this means that the the charge Q and the coulomb constant do  not change

  This means that

Where is the radius of the sphere

     and is the distance  from that point where the second potential was measured to the center of the sphere which is mathematically represented as

Substituting  this into the equation

 Now substituting value

From the equation above

making Q the subject

k has a values of

       Substituting into the equation

According to  Gauss law  the electric field from  outside a sphere is taken to be an electric field from a point charge this mean that the potential outside a sphere is also taken as electric potential inside a sphere

The magnitude of a electric field from a sphere (point charge ) is mathematically represented as

Substituting values

 According the the law of energy conservation

  The electric potential energy at the point outside the surface where the second potential was measured(21 cm from the sphere surface) = The electric potential energy at the surface + The kinetic energy of the charge (electron) at that the surface

Generally Electric potential energy is mathematically represented as

Where is e is an electron

And Kinetic energy is mathematically represented as

From the statement above

But from the question we can deduce that the potential at the surface is zero

So the equation becomes

The charge an electron has a value  

And the mass of an electron is

     Making v the subject

Substituting value

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer