The Electric Field at the Surface of the Sphere

The electric field at the surface of the sphere can be determined by using Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of the medium. In this case, we can consider a spherical Gaussian surface with radius R, centered at the center of the sphere.

Using Gauss's law, the electric flux through the Gaussian surface is given by:

Φ = E * 4πR²

Where E is the electric field at the surface of the sphere, and R is the radius of the Gaussian surface.

The charge enclosed within the Gaussian surface can be calculated by multiplying the volume charge density by the volume of the sphere:

Q = ρ * (4/3)πR³

Where ρ is the volume charge density.

Substituting the values of Φ and Q into Gauss's law equation, we get:

E * 4πR² = (ρ * (4/3)πR³) / ε₀

Where ε₀ is the permittivity of the medium.

Simplifying the equation, we find:

E = (ρ * R) / (3ε₀)

The Electric Potential at the Center of the Sphere

The electric potential at a point in an electric field is defined as the work done per unit charge to bring a positive test charge from infinity to that point.

Since the electric field is uniform within the sphere, the electric potential at the center of the sphere can be calculated using the formula:

V = - ∫ E · dr

Where V is the electric potential, E is the electric field, and dr is the displacement vector.

In this case, since the electric field is radial and the displacement vector is also radial, the dot product simplifies to:

V = - ∫ E dr

The integral can be evaluated as:

V = - ∫ E dr = - ∫ (ρ * R) / (3ε₀) dr

Since the electric field is constant, we can take it out of the integral:

V = - (ρ * R) / (3ε₀) ∫ dr

Evaluating the integral, we find:

V = - (ρ * R) / (3ε₀) * r

Where r is the distance from the center of the sphere to the point where the electric potential is being calculated.

Therefore, at the center of the sphere (r = 0), the electric potential will be zero.