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Q.
Flux passing through shaded surface of sphere when a point charge $q$ is placed at the centre is (radius of the sphere is $R$ )
Electric Charges and Fields
Solution:
$\alpha=60^{\circ}$
Solid angle subtended by $B C D$, $\Omega=2 \pi(1-\cos \alpha)=\pi$
Solid angle subtended by $A B D E$,
$\Omega_{(A B C D E)}-\Omega_{(B C D)}=2 \pi-\pi=\pi$
Hence, flux through $A B D E: \phi=\frac{q}{\varepsilon_{0}} \frac{\pi}{4 \pi}=\frac{q}{4 \varepsilon_{0}}$
