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
Answer & explanation
Correct answer: option 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]