Answer to Question #147168 in Electricity and Magnetism for Muhammad Moiz

Question

The nucleus of a plutonium-239 atom contains 94 protons.

Assume that the nucleus is a sphere with radius 6.64 fm and with

the charge of the protons uniformly spread through the sphere.At

the surface of the nucleus, what are the (a) magnitude and (b) direction

(radially inward or outward) of the electric field produced by

the protons?

Answer 1

• First we will an swer the question about direction : as we suppose that a nucleus is a uniformly charged sphere, by studying the symmetries of a problem we conclude that it is directed radially and the magnitude dependa only on R (distance from center of a nucleus). In addition, as proton charge is positive, it is directed outward.
• Now for the magnitude the question is to find the magnitude of an electric field of a uniformly charged sphere on it's surface. By using the Gauss theorem for a surface being a surface of a nucleus we can conclude that

$E(r) \times 4\pi r^2 = Q/\varepsilon_0 = 94\times |e|/\varepsilon_0$ , where $|e|$ is a charge of proton

and thus $E(r) = \frac{94|e|}{4\pi\varepsilon_0 r^2} = \frac{94 \cdot 1.6\cdot10^{-19}}{4\cdot \pi \cdot 8.85\cdot10^{-12}\cdot 6.64^2\cdot 10^{-30}} \approx 3.06 \cdot 10^{21} V/m$