Answer to Question #270768 in Physics for Betman

Question

The nucleus of an atom of uranium - 238 has a radius of 6.8×10^-15m and carries a positive charge of Ze in which Z(=92) is the atomic number of uranium and e is the elementary charge, (1) What is the electric field at the surface of such a nucleus?, (2) In what direction does it point?, (3) Does it's numerical value surprise you?, (4) Explain your answer.

Answer 1

$$E=\frac{(8.99\cdot10^{9})(92\cdot 1.6\cdot10^{-19})}{(6.8\cdot10^{-15})^2}=2.9\cdot10^{21}\frac{V}{m}$$

We have to calculate the net electric field at the location of the nucleus. The net electric field at any location inside or outside the nucleus or atom is the sum of the electric field produced by the nucleus and the electrons.

According to Gauss law in electrostatics, the electron cloud modeled as a spherical shell can cause no electric field at any point inside the spherical shell. Therefore electrons revolving around the nucleus do not produce an electric field at the location of the nucleus.

If we take the location of the nucleus as a point at the center of the nucleus, the net electric field at that point is also zero. Because we assumed the nucleus as a uniform solid sphere having volume charge density. The point at the center of the nucleus does not enclose any electric charge.

If we consider the location of the nucleus as a sphere of size equal to the size of the nucleus, the electric field at the location of the nucleus is equal to the electric field available at the surface of the nucleus.