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The charge distribution for this model of the atom is as shown in the above figure.

The total negative charge in the uniform spherical charge distribution of radius R must be –Ze, since the atom (nucleus of charge Ze + negative charge) is neutral.

This immediately gives us the negative charge density $\rho$, since we must have

$\large\frac{4 \pi R^3}{3} P=0 -ze $

$P=-\large\frac{-3 Ze}{4 \pi R^3}$

To find the electric field E(r) at a point P which is a distance r away from the nucleus, we use Gauss’s law.

Because of the spherical symmetry of the charge distribution, the magnitude of the electric field E(r) depends only on the radial distance, no matter what the direction of r.

. Its direction is along (or opposite to) the radius vector r from the origin to the point P.

The obvious Gaussian surface is a spherical surface centred at the nucleus. We consider two situations, namely, r < R and r > R.

(i) r < R : The electric flux $\phi$ enclosed by the spherical surface is

$\phi =E(r) \times 4 \pi r^2$

where E (r) is the magnitude of the electric field at r. This is because the field at any point on the spherical Gaussian surface has the same direction as the normal to the surface there, and has the same magnitude at all points on the surface.

The charge q enclosed by the Gaussian surface is the positive nuclear charge and the negative charge within the sphere of radius r,

i.e $q= Ze + \large\frac{4 \pi r^3}{3}$$p$

Substituting for the charge density obtained earlier, we have

The electric field is directed radially outward.

(ii) r > R: In this case, the total charge enclosed by the Gaussian spherical surface is zero since the atom is neutral. Thus, from Gauss’s law,

$E(r) \times 4 \pi r^2 =0$

$E(r) =0$

At $r = R,$ both cases give the same result: $E = 0.$

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