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.

 

[kyle8921]

First of all, it is important to clarify that the formula for charge, Q = VR, is only applicable for a capacitor, not a solid conduction sphere. For a solid sphere, the charge is not dependent on the voltage alone, but also on the material properties and geometry of the sphere. Therefore, the charge cannot be determined solely by the given voltage and radius.

As for the calculation of the electric field, your formula is correct. However, it is important to note that the electric field just outside the sphere is not only dependent on the charge, but also on the surrounding medium (represented by the permittivity, epsilon). So the value you have calculated may vary depending on the specific medium surrounding the sphere.

In conclusion, it is important to use the correct formulas and to consider all relevant factors when studying the behavior of a conduction sphere. It is also important to keep in mind that these calculations are simplified models and may not accurately represent real-world scenarios. Further experimentation and analysis would be necessary to fully understand the behavior of a conduction sphere.