Ncert Exercises & Solutions

1.29 A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is ( $\frac{σ}{2 ϵ_{0}} \hat{n}$ , whereis the unit vector in the outward normal direction, and σ 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, a is the charge density and $ε_{0}$ is the permittivity of free space.
Charge q = $σ$ X ds

According to Gauss's law, flux, $\emptyset = E \cdot d s = \frac{q}{ϵ_{0}}$

$\begin{array}{l} \Rightarrow E . d s = \frac{σ \times d s}{ϵ_{0}} \\ ∴ E = \frac{σ}{2 ϵ_{0}} \hat{n} \end{array}$

Therefore, the electric field just outside the conductor is $\frac{σ}{2 ϵ_{0}} \hat{n}$ . 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. $∴ E + E^{'} = E$

$\Rightarrow E^{'} = \frac{E}{2} = \frac{σ}{2 ϵ_{0}} \hat{n}$

Hence, the field due to the rest of the conductor is $\frac{σ}{ϵ_{0}} \hat{n}$ .

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