A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\left(\sigma / 2 \varepsilon_{0}\right) \hat{ n },$ where $\hat{ n }$ is the unit vector in the outward normal direction, and $\sigma$ is the surface charge density near the hole.
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\left(\sigma / 2 \varepsilon_{0}\right) \hat{ n },$ where $\hat{ n }$ is the unit vector in the outward normal direction, and $\sigma$ is the surface charge density near the hole.
Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero. Let $E$ is the electric field just outside the conductor, $q$ is the electric charge, $\sigma$ is the charge density and $\epsilon_{0}$ is the permittivity of free space. Charge $q=\sigma \times d s$
According to Gauss's law, flux, $\phi=E . d s=\frac{q}{\epsilon_{0}}$
$\Rightarrow E \cdot d s=\frac{\sigma \times d s}{\epsilon_{0}}$
$\therefore E =\frac{\sigma}{2 \epsilon_{0}} \hat{n}$
Therefore, the electric field just outside the conductor is $\frac{\sigma}{2 \epsilon_{0}} \hat{n} .$ This field is a superposition of field due to the cavity $E '$ and the field due to the rest of the charged conductor $E'$. These fields are equal and opposite inside the conductor and equal in magnitude and direction outside the conductor. $\therefore E '+ E '= E$
$\Rightarrow E'=\frac{ E }{2}=\frac{\sigma}{2 \epsilon_{0}} \hat{n}$
Hence, the field due to the rest of the conductor is $\frac{\sigma}{\epsilon_{0}} \hat{n}$
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- [JEE MAIN 2015]