The discussion focuses on the factors affecting the electric field around a charged conduction sphere, specifically one charged to 500 volts with a radius of 1 cm. The charge, calculated using the formula Q = VR, results in a value of 5. The electric field (E) just outside the sphere is derived from the equation E = (1/4πε) x (charge/R²), yielding an electric field strength of approximately 4.5 x 10^14 N/C. The participants emphasize the importance of unit consistency and the inverse relationship of the electric field strength to the square of the distance from the sphere.
PREREQUISITESPhysics students, electrical engineers, and anyone interested in understanding electrostatics and electric field behavior around charged objects.
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.Cairrd said: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?