What factors affect the electric field around a charged conduction sphere?

[Cairrd]

Ok for example, if a solid conduction sphere was charged to 500volts and had radius of 1cm. Then charge, Q = VR = 5?

So the E field just outside the sphere would be E = (1/4pi epsilon) x charge/R^2 = 4.5 x 10^14?

Or am i getting confused?

 

[OlderDan]

Ok for example, if a solid conduction sphere was charged to 500volts and had radius of 1cm. Then charge, Q = VR = 5?

So the E field just outside the sphere would be E = (1/4pi epsilon) x charge/R^2 = 4.5 x 10^14?

Or am i getting confused?

Just check your units. The E field anywhere outside the sphere is inversely proportional to r^2. You can find E just outside the surface, or anywhere else outside.