Answer:
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