•• A metal sphere carrying an evenly distributed charge will have spherical equipotential surfaces surrounding it. Suppose the sphere’s radius is 50.0 cm and it carries a total charge of (a) Calculate the potential of the sphere’s surface. (b)You want to draw equipotential surfaces at intervals of 500 V outside the sphere’s surface. Calculate the distance between the first and the second equipotential surfaces, and between the 20th and 21st equipotential surfaces. (c) What does the changing spacing of the surfaces tell you about the electric field?

Explanation:

For this exercise we will use that the potential is created by the charge inside the equinoctial surface and just like in Gauss's law we can consider all the charge concentrated in the center.

Therefore the potential on the ferric surface is

        V = k Q / r

where k is the Coulomb constant, Q the charge of the sphere and r the distance from the center to the point of interest

a) On the surface the potential

        V = 9 10⁹ Q / 0.5

        V = 18 10⁹ Q

Unfortunately you did not write the value of the load, suppose a value to complete the calculations Q = 1 10⁻⁷ C, with this value the potential on the surfaces V = 1800 V

b) The equipotential surfaces are concentric spheres, let's look for the radii for some potentials

for V = 1300V let's find the radius

             r = k Q / V

             r = 9 109 1 10-7 / 1300

             r = 0.69 m

other values are shown in the following table

V (V)      r (m)

1800     0.5

1300     0.69

 800      1,125

 300     3.0

In other words, we draw concentric spheres with these radii and each one has a potential difference of 500V

C) The spacing of the spheres corresponds to lines of radii of the electric field that have the shape

         E = k Q / r²

the answer is kevlar

the total distance traveled by the car is

30km + 49km = 79 km .