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1. Electric Charges and Fields
medium
In a cuboid of dimension $2 L \times 2 L \times L$, a charge $q$ is placed at the centre of the surface ' $S$ ' having area of $4 L ^2$. The flux through the opposite surface to ' $S$ ' is given by
A
$\frac{ q }{12 \varepsilon_0}$
B
$\frac{ q }{3 \varepsilon_0}$
C
$\frac{ q }{2 \varepsilon_0}$
D
$\frac{q}{6 \varepsilon_0}$
(JEE MAIN-2023)
Solution
$\phi=\frac{ Q / \varepsilon_0}{6}$
Flux passing through shaded face $=\frac{ q }{6 \varepsilon_0}$
Standard 12
Physics
