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

Now in order to find the electric field inside the tiny hole on the surface of a charged hollow conductor, we will assume the hole to be filled up with the same conductor with same charge density i.e. Hollow conductor without any hole and will find out electric field at the point of hole on the conductor due to the whole conductor and due to the hole element and thus finding the contribution in electric field due the hole element and subtracting it with the total electric field of the conductor without any hole we will get the electric field of the rest of the conductor with hole.

Now consider a hollow spherical conductor with surface charge density C/m² with a tiny on its surface now considering two points one just outside the surface of conductor near the hole say A and other inside the hollow cavity say B

As shown in the figure

Now finding the electric field at point A due to the whole conductor assuming hole to be filled up now the total charge on the conductor with total surface area S and surface charge density will be

q = × S

From Gauss Theorem we know, net electric flux through an Gaussian surface enclosing a charge is equal to net charge enclosed by surface divided by permittivity of free space

Mathematically:

Where E is the electric field so denotes the net Electric flux through the surface q is the net charge enclosed by the Gaussian surface and is permittivity of free space which is a constant

Now assuming an Gaussian surface just enclosing the conductor

As shown in the figure

So we will have

Since electric field is always normal to a conductor surface so E and ds are in same direction i.e. = 0

So, cos(θ) = 1 and due to spherical symmetry magnitude of electric field will same everywhere so we have

= E × S = q/

i.e. E × S = ( × S) /

or E = /ϵ₀

now since electric field is always normal to the surface of a charged conductor we have

E = (/ϵ₀)

Where ń is the outward normal to the surface of conductor

Now let us assume contribution of the tiny hole to this electric field is E₁ and Electric field due to rest of the conductor is E₂

So, we have E = E₁ + E₂ (Just outside the conductor)

Now inside the hollow cavity inside the conductor at point B we know electric field is zero because of electrostatic shielding

Again assuming the electric field at that point to be contribution of the tiny hole to this electric field is E₁ and Electric field due to rest of the conductor is E₂ now since the hole will act as a point sized charge so field would be directed outwards i.e. in vertically downward direction, and rest of the conductor is below the point so field would be in vertically upward direction, i.e. field due to both hole and rest of conductor are opposite in direction and net field is zero inside a cavity in a conductor

So, we have,

0 = E₁ - E₂ (Just inside the hollow cavity of conductor)

Or E₁ = E₂

i.e. Electric field due to hole and rest of conductor are equal in magnitude and in same direction outside the conductor and equal in magnitude and opposite in direction in the hollow cavity.

Now using, E₁ = E₂ and E = E₁ + E₂

We have, 2E₁ = E and 2E₂ = E

Or, E₁ = E/2 and E₂ = E/2

Also we know E = (/)

i.e. the electric field due to the hole only is

E₁ = (/2)

And electric field due to rest of surface when hole is made is also

E₂ = (/2)

i.e. the electric field in the hole of a charged hollow conductor is (/2)

Where is a unit vector in the outward normal direction.