We find that the electric field of a charged disk approaches that of a charged particle for distances y that are large compared to R, the radius of the disk. To see a numerical instance of this, calculate the magnitude of the electric field a distance y = 2.9 m from a disk of radius R = 2.9 cm that has a total charge of 5.0 µC using the exact formula as follows. (Enter your answer to at least one decimal place.)
E = 2πK σ(1- [y/(R² + y²)]) j^

You are a researcher for a golf club manufacturer. you are given two identical looking cubes of a metal alloy. you are informed that they are made of the exact same material, but one is crystalline, whereas the other is amorphous. it is your job to determine which one is amorphous this one is more stress-resistant and is useful in reinforcing golf clubs. which of the following is the best way to determine which is which? a. determine the density of each cube. the more dense one is the amorphous solid. b. melt both cubes and look for a broader range of melting temperatures. the one that melts over a broader range of temperatures is the amorphous solid. c. determine the density of each cube. the less dense cube is the amorphous solid. d. melt both cubes and measure the range of melting temperatures. the one that melts over a narrower range of temperatures is the amorphous solid.