If a small piece of radius b is removed from a charge spherical shell of radius a (>> b), the electric intensity at the mid-point of the aperture, assuming the density of charge to be σ, will be

$\frac{\sigma}{2 \varepsilon_0}$

We know that for a charged spherical shell (or conductor)

$E_{out}=\frac{\sigma}{\varepsilon_0}$ and $E_{\text {in }}=0$           …(i)

Let E_D and E_R be the intensities due to disc and the remainder respectively at P. As shown in figure, for outside the shell both E_D and E_R will be directed outwards, but for inside, E_R will be outwards and E_D inwards, so that

E_out = E_R + E_D

and E_in = E_R − E_D         …(ii)

And hence from Eqns. (i) and (ii),

$E_R+E_D=\frac{\sigma}{\varepsilon_0} \quad \text { and } \quad E_R-E_D=0$

Solving these for $E_R$ and $E_D$, we get

$E_R=E_D=\frac{\sigma}{2 \varepsilon_0}$

Thus, the field at the aperture will be (σ/2ε₀) directed outwards.

∴ (b)