First slide

Gauss's Law

Question

An early model for an atom considered to have a positively charged point nucleus of charge Z e surrounded by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, what is the electric field at a distance r from the nucleus, where r>R ?

Moderate

A

$\frac{1}{4 {πε}_{0}} \frac{Ze}{r^{2}}$
B

$\frac{1}{4 {πε}_{0}} \frac{Ze}{R^{2}}$
C

$\frac{1}{4 {πε}_{0}} \frac{Ze}{r^{2} - R^{2}}$
D

0

Solution

The situation of charge distribution in the given model is depicted in the following figure:

Let p be the point (whose distance is r from positively charged point nucleus of charge Ze) at which the electric field needs to be measured. The Gaussian sphere is drawn through point P.
Due to the spherical symmetry of the charge distribution, the magnitude of the electric field (E) of the electric field the same at all points of the Gaussian sphere, which acts radially outwards (due to +Z e ) and acts radially inwards (due to -Z e). Now, the electric flux due to Gaussian surface is given by
$ϕ_{E} = E (4 {πr}^{2})$ ….(i)
We know that according to Gauss's theorem, $ϕ_{E} = q / ε_{0}$ , where q is the net charge enclosed by the Gaussian sphere (through point P, in this case). Since the atom is neutral, q=0, Therefore, from Eq. (1), we get
and hence the electric field outside the atom (where r>R) is zero.

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