Answer:
Consider a uniformly charged spherical shell of radius R as shown. Let \[\sigma \] be the uniform surface charge density of the shell. find the electric field at a point P outside the shell we take the Gaussian surface to be a sphere of radius r and centre O passing through P. The electric field at each point of the Gaussian surface has the same magnitude E and is along the radius vector at each point. If we consider a small elemental surface dS at P, the flux through ds is\[d\phi =E\cdot ds=Eds\cos 0=Eds\] \[\therefore \] The total flux through the Gaussian surface is \[E\times 4\pi {{r}^{2}}\] The charge enclosed is \[\sigma \times 4\pi {{R}^{2}}.\] According to Gauss theorem, the electric flux through a closed surface S, \[{{\phi }_{E}}=\frac{q}{{{\varepsilon }_{0}}}\] where, q is the total charge enclosed by S and \[{{\varepsilon }_{0}}\] is the permeability in free space. By Gauss?s law, electric flux \[=\frac{q}{{{\varepsilon }_{0}}}\] Substituting the values, we get \[E\times 4\pi {{r}^{2}}=\frac{\sigma \times 4\pi {{R}^{2}}}{{{\varepsilon }_{0}}}\] i.e., \[E=\frac{\sigma {{R}^{2}}}{{{\varepsilon }_{0}}{{r}^{2}}}=\frac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] where, \[q=4\pi {{R}^{2}}\sigma \] is the total charge on the spherical shell. But \[\frac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] is the electric field produced by a charge q placed at the centre O. Hence, 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.
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