A hemisphere of radius R is placed between two uniformly charged-thin infinite sheets of charge density $+σ$ and $-σ$. Flux through the curved surface of the hemisphere

$\frac{σ}{ε_0}(πR^2)$

The correct answer is Option (2) - $\frac{σ}{ε_0}(πR^2)$

$\phi_ε=\int\vec E.d\vec a$

where,

$\phi_ε$ = flux

$\vec E$ = Electric field intensity

$d\vec a$ = Area vector

$\vec E$ produced by thin infinite sheet = $\frac{σ}{ε_0}$ [formula]

$∴\phi_ε=\int\frac{σ}{ε_0}.da$

$=\frac{σ}{ε_0}\int da=\frac{σ}{ε_0}(\pi R^2)$

[As θ = 0°, because $\vec E$ are perpendicular to the base of hemisphere]