What is the electric potential on the surface of a conducting sphere of radius R carrying charge Q in vacuum?

What is the electric potential on the surface of a conducting sphere of radius R carrying charge Q in vacuum?

• A. V_surface = (1/(4π ε0)) (R / Q).
• B. V_surface = (1/(4π ε0)) (Q / R).
• C. V_surface = (1/(4π ε0)) (Q / (2R)).
• D. V_surface = (1/(4π ε0)) (Q R).

Answer: (B)

When a conductor is in vacuum, all the charge sits on the surface and the electric field inside is zero, so the potential is constant everywhere inside and on the surface. Outside the sphere, the field looks like that of a point charge Q at the center: E = (1/4π ε0) Q / r^2 directed radially outward. To find the surface potential, take the reference V(∞) = 0 and integrate the field from infinity to the surface: V(R) = -∫∞^R E·dr = -∫∞^R (1/4π ε0) Q / r^2 dr = (1/4π ε0) Q / R. Thus the electric potential on the surface is (1/4π ε0) (Q / R).

When a conductor is in vacuum, all the charge sits on the surface and the electric field inside is zero, so the potential is constant everywhere inside and on the surface. Outside the sphere, the field looks like that of a point charge Q at the center: E = (1/4π ε0) Q / r^2 directed radially outward. To find the surface potential, take the reference V(∞) = 0 and integrate the field from infinity to the surface:

V(R) = -∫∞^R E·dr = -∫∞^R (1/4π ε0) Q / r^2 dr = (1/4π ε0) Q / R.

Thus the electric potential on the surface is (1/4π ε0) (Q / R).