1.29 A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (<math><mfrac><mrow><mrow><mi mathvariant="bold-italic">σ</mi></mrow></mrow><mrow><mrow><mn>2</mn><msub><mrow><mrow><mi mathvariant="bold-italic">ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac></math> ) , where <math><mover accent="true"><mrow><mrow><mi mathvariant="bold-italic">n</mi></mrow></mrow><mo>̂</mo></mover></math> is the unit vector in the outward normal direction, and <math><mi mathvariant="bold-italic">σ</mi><mi mathvariant="bold-italic"></mi></math> is the surface charge density near the hole.

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Ncert Solutions Physics Class 12th Physics Ncert Solutions Class 12th Physics Class 12th Electric Charges and Fields

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

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alok kumar singh | Contributor-Level 10

3 months ago

1.29 Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero. Let E be the electric field outside the conductor, q is the electric charge, $σ$ is the charge density and $ε_{0}$ is the permittivity of free space.

Charge q = $σ$ $\times d s$

According to Gauss’s law, flux $φ$ = E.ds = $\frac{q}{?_{0}}$ = $\frac{σ \times d s}{?_{0}}$

Hence, E = $\frac{σ}{2 ε_{0}} n$

Therefore, the electric field just outside the conductor is $\frac{σ}{{2 ε}_{0}} n$ . 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.

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Charge q = $σ$ $\times d s$

According to Gauss’s law, flux $φ$ = E.ds = $\frac{q}{?_{0}}$ = $\frac{σ \times d s}{?_{0}}$

Hence, E = $\frac{σ}{2 ε_{0}} n$

So E’ + E’ = E

E’ = $\frac{E}{2}$ = $\frac{σ}{2 ε_{0}} n$

Hence, the field due to the rest of the conductor is $\frac{σ}{2 ε_{0}} n$
$?$

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1.29 Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero. Let E be the electric field outside the conductor, q is the electric charge, <math><mi>σ</mi></math>is the charge density and <math><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></math> is the permittivity of free space.Charge q =<math><mi>σ</mi></math> <math><mo>×</mo><mi>d</mi><mi>s</mi></math> According to Gauss’s law, flux<math><mi>φ</mi></math> = E.ds = <math><mfrac><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><msub><mrow><mrow><mi>?</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac></math>=<math><mfrac><mrow><mrow><mi>σ</mi><mi mathvariant="normal"></mi><mo>×</mo><mi>d</mi><mi>s</mi></mrow></mrow><mrow><mrow><msub><mrow><mrow><mi>?</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac></math>Hence, E = <math><mfrac><mrow><mrow><mi>σ</mi></mrow></mrow><mrow><mrow><mn>2</mn><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac><mover accent="true"><mrow><mrow><mi>n</mi></mrow></mrow></mover></math>Therefore, the electric field just outside the conductor is <math><mfrac><mrow><mrow><mi>σ</mi></mrow></mrow><mrow><mrow><msub><mrow><mrow><mn>2</mn><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac><mover accent="true"><mrow><mrow><mi>n</mi></mrow></mrow></mover></math> . 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.So E’ + E’ = EE’  = <math><mfrac><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></math>=<math><mfrac><mrow><mrow><mi>σ</mi></mrow></mrow><mrow><mrow><mn>2</mn><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac><mover accent="true"><mrow><mrow><mi>n</mi></mrow></mrow></mover></math>Hence, the field due to the rest of the conductor is <math><mfrac><mrow><mrow><mi>σ</mi></mrow></mrow><mrow><mrow><mn>2</mn><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac><mover accent="true"><mrow><mrow><mi>n</mi></mrow></mrow></mover></math> <math><mfrac><mrow><mrow><mi></mi></mrow></mrow></mfrac><mover accent="true"><mo>?</mo></mover></math>

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

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1.29 A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ2ε0 ) , where n̂ is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.

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