A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is n(σ2∈0)n^. where nn^ is the unit vector in - Physics

Numerical

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (sigma/(2in_0)) hat"n". where hat"n" is the unit vector in the outward normal direction, and sigma is the surface charge density near the hole.

Solution

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, sigma is the charge density, and in_0 is the permittivity of free space.

Charge |"q"| = vecsigma xx vec("ds")

According to Gauss’s law,

Flux, phi = vec"E".vec("ds") = |"q"|/in_0

"E"."ds" = (vecsigma xx vec("ds"))/in_0

∴ "E" = sigma/in_0.hat"n"

Therefore, the electric field just outside the conductor is sigma/in_0.hat"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' = "E"/2

= sigma/(2in_0).hat"n"

Therefore, the field due to the rest of the conductor is sigma/in_0.hat"n".

Hence, proved.

Concept: Continuous Distribution of Charges
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APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 1 Electric Charges and Fields
Exercise | Q 1.29 | Page 49
NCERT Class 12 Physics Textbook
Chapter 1 Electric Charge and Fields
Exercise | Q 29 | Page 49

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