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# A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/2ε0 ) nˆ , where nˆ is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.

Q 1.29: A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\left (\frac{\sigma}{2 \epsilon_{0}} \right )$ $\widehat{n}$ , where $\widehat{n}$ is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.

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Let the surface area of the sphere be S.

And assume that the hole is filled. For a point B, just above the hole, considering a gaussian surface passing through B, we have

$\oint E.dS = q/\epsilon_{0}$

Now, Since the electric field is always perpendicular to the surface of the conductor.

$\\ \therefore \oint E.dS = E.S= q/\epsilon_{0} = (\sigma.S)/\epsilon_{0} \\ \implies E = \sigma/\epsilon_{0}$

Using Superposition principle,  $E =E_{1} + E_{2}$ ,

where $E_{1}$ is due to the hole and  $E_{2}$ is due to the rest of the conductor. (Both pointing outwards, i.e away from the centre)

Again, for a point A, just below the hole, Electric field will be zero because of electrostatic shielding.

Using the superposition principle, this will be due to $E_{1}$  pointing inwards(towards the centre) and due to $E_{2}$(Pointing away from the centre)

$0 =E_{1}-E_{2}$  $\implies E_{1}=E_{2}$

Using this relation, we get:

$E =E_{1} + E_{2} = 2E_{1} \implies E_{1} = E/2 = \sigma/2\epsilon_{0}$

Since this is pointing outwards,

$\boldsymbol{E_{1}} = \sigma/2\epsilon_{0}\ \boldsymbol{\widehat{n}}$   is the electric field in the hole.

(Trick: 1. Assume the hole to be filled.

2. Consider 2 points just above and below the hole.

3. Electric fields at these points will be due to the hole and rest of the conductor. Use superposition principle.)

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