Question

An infinitely large non conducting plane sheet of charge density sigma has a circular aperture of certain radius carved out of it the electric field at a point which is at distance r from the centre of aperture is sigma bond and not ready for that person is

Answer 1

Answer: r = a

Remark, Question is the relation between r and a.

If effective electric field intensity given.

Explanation: σ/2ε

Electric field intensity  at a distance r.

σ/2ε*$(1-\frac{r}{\sqrt{a^{2}+r^{2}  } })$

Carved portion removed,

Hence, replace σ = -σ.

-σ/2ε*$(1-\frac{r}{\sqrt{a^{2}+r^{2}  } })$.

Effective electric intensity.

σ/2ε+-σ/2ε*$(\frac{1}{\sqrt{a^{2}+r^{2}  } })$

σr*$(1-\frac{r}{\sqrt{a^{2}+r^{2}  } })$ = σ/2ε*($\frac{1}{\sqrt{2} }$)

r= a.

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