we know that electric field of an ideal spherical shell with uniformly distributed charge is zero inside the shell and equal to EF of a point charge on its center.
when we calculate the EF for a point on the surface of shell,it is equal to half of EF of same point charge EF.
all the above statements are based on hypothesis that charge is distributed continuously and there are infinite small charges.
but here the question is that what would be the EF on surface by doing experiment or a more realistic simulation.
The Gauss' theorem and the spherical symmetry are bound to answer completely this question. Gauss' theorem says that if we consider a closed surface S around a charge Q, then the integral over the surface of the electric field E is equal to the charge inside the surface
∫SE dS = Q/ε
where E is the electric field vector, dS is a vector along the normal to the surface element dS, and ε is the electric permittivity inside the closed surface S.
Thus, if we choose the surface S spherical, of radius R, and centered on the center of the charge Q, the spherical symmetry imposes E to be of the same absolute value all over the surface S, and oriented along the radius. Therefore we have
ES = Q/ε ==> 4πR2E = Q/ε ==> E = Q/(4πR2ε).
The reality is that the shell is not perfectly spherical and charges distribution will depends on the shape and not be uniformly.
A modified Brown and Hemingway's method (Eq.2-6) can be applied to calculate the spatial charge distribution of an arbitrarily shaped particles. The appendix 1 helps you solve Eq.2-6 and appendix 2 helps you break a particle into thousands of small elements.
With the spatial charge distribution, you can then calculate the EF.
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(Oct 7, 2020): Let us consider the model of a conducting spherical shell with radius R containing a total amount of charge Q that is spread uniformly over its surface. This results in a constant uniform surface charge density. By applying Gauss’s law one finds that:
a) Electric field inside the spherical shell at radial distance r from the center of the spherical shell so that rR)=k*Q/r^2 (k is Coulomb’s electric constant).
Special cases: On the surface (just outside): E(r=R+)=k*Q/R^2.
On the surface (just inside): E(r=R-)=0.
Uniform charge distribution represents also the static electrostatic equilibrium for a spherical shell in the sense that the spherical shell surface (and inside) is an equipotential surface.
If you are doing an experiment and there are charges around the spherical shell, either inside or outside the spherical shell, the electric field will probably not be zero inside the shell. What we can say, however, is that the electric field inside a spherical shell of charge is the same whether or not the shell is there. The electric field outside the shell will be much more complicated and its configuration will depend on the specific nature of the charges around the spherical shell
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