Question 1.29) A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ 2ε0) n , where n is the unit vector in the outward normal direct

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,

σ

is the charge density, and

ε_{ο}

is the permittivity of free space.

Therefore, the electric field just outside the conductor is

E = \frac{σ}{ε_{ο}} ˆ 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.

Therefore, the field due to the rest of the conductor is

= \frac{σ}{2 ε_{ο}} n ˆ

Question 1.29) A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ 2ε0) n , where n is the unit vector in the outward normal direct- Hosting Precise

Posted on 9th Oct 2021 09:15:20 in

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

Ans:

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, $σ$ is the charge density, and $ε_{ο}$ is the permittivity of free space.

Charge $|q| = \underset{σ}{\to} \times \underset{d s}{\to}$

According to Gauss’s law,

$Flux ϕ = \underset{E}{\to} . \underset{ds}{\to} = \frac{|q|}{ε_{ο}}$

$Eds = \frac{\underset{σ}{\to} \times \underset{ds}{\to}}{ε_{ο}}$

$E = \frac{σ}{ε_{ο}} ˆ n$

Therefore, the electric field just outside the conductor is $E = \frac{σ}{ε_{ο}} ˆ n$ . 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.

$E' + E' = E$

$E' = \frac{E}{2}$

$= \frac{σ}{2 ε_{ο}} n ˆ$

Therefore, the field due to the rest of the conductor is $= \frac{σ}{2 ε_{ο}} n ˆ$

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