if q > and inward if q < . This, however, is exactly the field produced by a charge q placed at the centre O. Thus for points outside the shell, the field due to a uniformly charged shell is as if the entire charge of the shell is concentrated at its centre. (ii) Field inside the shell: In Fig.
. (b), the point P is inside the shell. The Gaussian surface is again a sphere through P centred at O. FIGURE .
Gaussian surfaces for a point with (a) r > R , (b) r < R . E XAMPLE . The flux through the Gaussian surface, calculated as before, is E × π r . However, in this case, the Gaussian surface encloses no charge.
Gauss’s law then gives E × π r = i.e., E = ( r < R ) ( . ) that is, the field due to a uniformly charged thin shell is zero at all points inside the shell * . This important result is a direct consequence of Gauss’s law which follows from Coulomb’s law. The experimental verification of this result confirms the / r dependence in Coulomb’s law.
Example . An early model for an atom considered it to have a positively charged point nucleus of charge Ze , 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?
FIGURE . Solution The charge distribution for this model of the atom is as shown in Fig. . .
The total negative charge in the uniform spherical charge distribution of radius R must be – Z e , since the atom (nucleus of charge Z e + negative charge) is neutral. This immediately gives us the negative charge density ρ , since we must have – Ze ρ or Ze ρ = − To find the electric field E ( r ) at a point P which