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,

σis the charge density andεis the permittivity of free space._{0}Charge

q = σ x dsAccording to Gauss’s law, flux,

φ = E.ds = q/ε_{0}⇒

E.ds=(σ x ds)/ε_{0}Therefore

E= (σ/2ε_{0}) ñTherefore, the electric field just outside the conductor is

(σ/2ε. This field is a superposition of field due to the cavity_{0}) ñÉand 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/2ε_{0}) ñHence, the field due to the rest of the conductor is (σ/ε

_{0}) ñ .