A solid sphere of radius R has a constant charge distribution, r, throughout its volume. Calculate the electric potential, V(r =0), at the center of the sphere assuming that infinitely far away V(r =[infinity]) = 0.

  Vb = k Q / 2R

Explanation:

To calculate the electric potential we use the relationship with the electric potential

         dV = E. ds

Therefore we must find the electric potential for the interior of the sphere, for which we can use Gauss's law.

         Ф = E .dA = / ε₀

We define a Gaussian surface as a sphere of radius r, in this case the electric field lines and the radial lines of the sphere are parallel so the scalar product is reduced to the algebraic product

        The area to a sphere is

         A = 4π R²

         E A = q_{int} /ε₀

Let's use constant charge density

             ρ = q_{int} / V

             V = 4/3 π r³

            q_{int} = ρ 4/3 π r³

We substitute

             E = (ρ 4/3 π r³) / 4π r² ε₀

             E = ρ / 3ε₀ r

Now we can find electrical potential

             .dV = - ρ / 3ε₀ ∫ r dr

             Vb –Va = - ρ / 3ε₀ r² / 2

We evaluate between the lower limits r = R and the upper limit r = r

           Vb –Va = ρ / 3ε₀  (R² - r²) / 2

We substitute the density value

             ρ = Q / (4/3 π R³)

          Vb- Va = Q (1 / 4πε₀) (R²-r²) / 2R³

  Where

           k = 1 / 4π ε₀

          Vb-Va = k Q / 2R³ (R² - r²)

In the center of the sphere r = 0

          Vb = k Q / 2R

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

dang that balls went far