Simple Voltage in Conducting Spherical Shell Question

[Switch11b]

When working on homework problems having to do with voltage, there is often the stipulation that V=0 at infinite. If you are dealing with a spherical conductive shell with charge Q, the Electric Field E is equal to K*Q/R^2, where K is the vacuum permittivity constant and R is the radius of the shell. To find the voltage on the surface of the shell you would then use V=E*R, which equates to the general form of the voltage equation V=K*Q/R. Following this course of thought, the voltage in the center of the shell would be V=K*Q/R, where R is 0. Making the voltage in the center of the shell zero. Is the voltage in the center of a spherical conducting shell really zero or did my logic fail me somewhere?
Thanks

 

[kanato]

Inside the conducting shell, the voltage will be constant. Inside the shell of charge the E field is no longer K*Q/R^2. If you have a shell, the E field will be zero inside the shell, easily proved with Gauss's law. This means that the voltage will be constant inside the shell, and the voltage has to be continuous across the boundary of the shell so the constant value inside the shell will be equal to the potential at the surface of the shell.

 

[astrokreng]

for your question! Your logic is actually correct. In the case of a conducting spherical shell, the electric field and voltage are both zero at the center of the shell. This is because the charges on the shell are distributed evenly on the surface, resulting in a cancellation of electric field and voltage at the center. This phenomenon is known as "electrostatic shielding" and is a fundamental concept in electromagnetism. So, your understanding is correct and there is no error in your logic. Keep up the good work!