CBSE Class 12-science Answered

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