The electric field at the centre of a uniformly charged hemispherical shell is E₀. Now two portions of the hemisphere are cut from either side and remaining portion is shown in figure. If $α = β = \frac{π}{3}$ , then electric field intensity at centre due to remaining portion is

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$\frac{E_{0}}{6}$

$\frac{E_{0}}{2}$

$\frac{E_{0}}{3}$

E₀

answer is C.

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Detailed Solution

Consider the whole hemisphere as three portion if electric field due to one portion is E₁ then
2E₁ sin 30 + E₁ = E₀
2E₁=E₀
$\Rightarrow E_{1} = \frac{E_{0}}{2}$

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