Using Gauss's Law and the relation between electric potential and electric field, show that the potential outside a uniformly charged sphere is identical to the potential of a point charge placed at the center of the sphere and equal to the total charge of the sphere. What is the potential at the surface of the sphere? How does the potential change if the charge distribution is not uniform but has spherical (radial) symmetry?

Using Gauss's Law, which states that the total electric flux through a closed surface is equal to the charge enclosed divided by the dielectric constant, we find the electric field outside a uniformly charged sphere. For a sphere with radius R and total charge Q, the electric field E outside the sphere can be determined using the following equation: E = (1 / (4 * \pi * \epsilon_0)) * (Q / r^2) Here, E is the electric field, \epsilon_0 is the vacuum permittivity, r is the distance from the center of the sphere to the point where the electric field is being measured, and Q is the total charge of the sphere.

To find the electric potential outside the sphere, we use the relation between electric potential V and the electric field E: V = - \int E dr For our case, we'll find the potential outside the sphere: V = - \int (1 / (4 * \pi * \epsilon_0)) * (Q / r^2) dr Integrating the electric field equation from infinity to r gives: V = (1 / (4 * \pi * \epsilon_0)) * (Q / r)

The electric potential of a point charge Q is also given by: V = (1 / (4 * \pi * \epsilon_0)) * (Q / r) We can see that the electric potential outside a uniformly charged sphere is equal to the potential of a point charge placed at the center of the sphere.

To find the potential at the surface of the sphere, we replace r with R, the radius of the sphere: V_surface = (1 / (4 * \pi * \epsilon_0)) * (Q / R)

If the charge distribution is not uniform but has spherical (radial) symmetry, the electric field outside the sphere will still follow the same formula: E = (1 / (4 * \pi * \epsilon_0)) * (Q / r^2) The electric potential formula outside the sphere remains the same: V = (1 / (4 * \pi * \epsilon_0)) * (Q / r) However, the potential at the surface and the electric field inside the sphere will be different and would depend on the specific charge distribution. In conclusion, the potential outside a uniformly charged sphere is identical to the potential of a point charge placed at the center of the sphere, equal to the total charge of the sphere. The potential at the surface of the sphere is given by V_surface = (1 / (4 * \pi * \epsilon_0)) * (Q / R). If the charge distribution is not uniform but has spherical symmetry, the potential outside the sphere remains the same, but the potential at the surface and the electric field inside the sphere will be different.