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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 point A, just below the hole, the 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}}   it 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 the rest of the conductor. Use the superposition principle.)

Posted by

HARSH KANKARIA

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