A hollow charged conductor has a tiny hole cut into its surface. What will be the electric field in the hole?

Asked by nupur | 23rd Feb, 2011, 12:00: AM

Expert Answer:

Dear student,

consider the case as a superimposition of fields.

field due to charge at the position of the hole when it was not present(E1 radially outwards)

and field due to the rest of the part(E2 radially outwards)

when the hole was not present at all if we consider the field intensity at a point just inside the

position of the hole the field intensity at that point is zero(from gauss' theorem)

so -E1+E2=0 (the point is just inside E1 is radially inwards hence the negative sign) 

just outside E=E1+E2=2E2=0

E2=0       radially outwards

Hope this helps.

Answered by  | 1st Mar, 2011, 10:38: AM

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