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# By the gauss’s theorem show that the electric field in a hollow spherical conductor is zero.

Last updated date: 22nd Jun 2024
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Since we have to find the electric field inside the hollow spherical conductor, thus the Gaussian surface would be a concentric hollow sphere whose radius, $r$ is less than the radius of the spherical conductor.
That is, if $R$is the radius of the given spherical conductor, then $rThe elemental area,$dA=4\pi {{r}^{2}}.dr$. The Gauss theorem is mathematically expressed as:$\oint{\vec{E}.d\vec{A}=\dfrac{{{q}_{enclosed}}}{{{\varepsilon }_{0}}}}$Where,$\vec{E}=$electric field of Gaussian surface.${{q}_{enclosed}}=$charge enclosed inside the gaussian surface${{\varepsilon }_{0}}=$permittivity of free space or vacuum We know that if we provide charge to a conductor, that charge resides only on the outer walls of the conductor and thus, there is zero charge inside the conductor. Now, we have been given a hollow spherical conductor. In this case as well, the charge will reside on the outer wall of the conductor and there is zero charge anywhere inside the conductor.$\Rightarrow {{q}_{enclosed}}=0$Thus, applying Gauss theorem, we have$\oint{\vec{E}.d\vec{A}=\dfrac{0}{{{\varepsilon }_{0}}}}\Rightarrow \vec{E}=0\$