# Apply Gauss law to show that for a charged spherical shell, the electric field outside the shell is, as if the entire charge were concentrated at the centre.

Let us consider a spherical shell P with centre 'O' and Radius (R), also consider a gaussian surface outside the shell at a distance r from the centre of the shell.

the electric flux through the surface

$\phi = E \times 4\pi r^{2}\; \; \; \; \; \; \; (i)$

from gauss law, we have

$\phi = \frac{q}{\varepsilon _{o}}\; \; \; \; \; \; \; (ii)$

Now, from equation (i) and (ii), we have

$\frac{q}{\varepsilon _{o}}=E\times 4\pi r^{2}$

$E=\frac{1}{4\pi \varepsilon _{o}}\frac{q}{r^{2}}$

This electric field is similar to that of a point charge q placed at the centre.

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