Class-12th Physics, CBSE and UP Board
Electric Charges and Fields,
Chapter-1, Exercise -1, Additional Exercise, Q-1.29
NCERT Solutions for Class 12th Physics
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.
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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 ε0 is the permittivity of free space.
Charge q = σ x ds
According to Gauss’s law, flux, φ = E.ds = q/ε0
⇒E.ds=(σ x ds)/ε0
Therefore E= (σ/2ε0) ñ
Therefore, the electric field just outside the conductor is (σ/2ε0) ñ . This field is a superposition of field due to the cavity É and the field due to the rest of the charged conductor É .These fields are equal and opposite inside the conductor and equal in magnitude and direction outside the conductor. Therefore É +É = E
⇒ É =E/2 = (σ/2ε0) ñ
Hence, the field due to the rest of the conductor is (σ/ε0) ñ .