A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (sigma)/( in_(0))hat(n), where hat(n) is the unit vector in the outward direction, and sigma is the surface density of charge near the hole .

A hollow charged conductor has a tiny hole cut into its surface. Show

1.	A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (sigma)/( in_(0))hat(n), where hat(n) is the unit vector in the outward direction, and sigma is the surface density of charge near the hole .
Answer» Solution :Suppose the hole has been plugged. Then the field inside the conductor is zero and outside it is `(sigma)/( in_(0)) hat (n)`. This FIELDIS essential the sum of two fields. One DUE to the plugged hole and other due to the RESET of the conductor. Inside the conductor these two forces are equal and opposite and the resultant is zero. But on the outer SURFACE these two forces are equal act in the same direction. Since the resultant is `( sigma)/( in_(0) ) hat(n)`, then each of these FORCE is equal to `( sigma)/( 2in_(0))hat (n) `. Thus the electric field in the hole is equal to `( sigma)/( in_(0)) hat (n)`.

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A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (sigma)/( in_(0))hat(n), where hat(n) is the unit vector in the outward direction, and sigma is the surface density of charge near the hole .

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