• # question_answer Using Gauss' theorem, show mathematically that for any point outside the shell, the field due to a uniformly charged spherical shell is same as if the entire charge is concentrated at the centre.

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.