The radius of a gold nucleus (Z=79) is about `7.0*10^(-10) m`. Assume that the positive charge is distributed uniformly throughout the nuclear volume. Find the strength of the electric field at (a) the surface of the nucleus and (b) at the middle point of a radius. Remembering that gold is a conductor, is it justified to assume that the positive charge is uniformly distributed over the entire volume of the nucleus and does not come to the outer surface?

Explanation

Charge present in gold nucleus
`=79 xx 1.6 xx (10^-19)C`
Since the surface enclose all the charges,
we have
(a) `E xx ds = Q/epsilon_0 `
` E = Q/epsilon_0 ds = x [area=4pi r^2]`
= ` 2.315131 xx (10^21)N/C.`
` (b) For the middle point of the radius, `
` Now, here r= 7/2 xx (10^15) m `
` volume = 4/3 pi r^3 `
` = 4/3 xx 22/7 xx 348/8 xx (10^-45)`
` Net charge = 7.9 xx 1.6 xx (10^-19)C `
Volume charge density
` = (7.9 xx 1.6 xx (10^-19)/ 4/3 pi xx 343 xx (10^-45))`
So charge m required part
` = (79 xx 1.6 xx (10^-19)/ 4/3 pi xx 343 xx (10^-45)) xx 4/3 pi xx 343/8 xx (10^-45)`
` = (79 xx 1.6 xx (10^-19)/8)`.
So, E = (79 xx 1.6 xx (10^-19)/4 pi (epsilon_0) .(r^2))`
` = 1.16 xx (10^21) N/C ` .