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  2. 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.
# NCERT

 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.

Answers (1)
H Harsh Kankaria

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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