A point charge + Q is placed just outside an imaginary hemispherical surface of radius R as shown in the…

Every point on circumference of flat surface is at equal distance from point charge 

Hence circumference is equipotential.

Flux passing through curved surface = - flux passing through flat surface.

${\left(d\varphi \right)}_{\mathrm{t}\mathrm{h}\mathrm{r}\mathrm{o}\mathrm{u}\mathrm{g}\mathrm{h}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{ }\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}}=E\cos \theta . dA=\frac{1}{4\pi {\in }_0} \frac{Q}{\left(\sqrt{r^2+R^2}\right)}\frac{R}{\sqrt{R^2+r^2}} . 2\pi rdr$

$\therefore \int d\varphi =\frac{QR}{4\pi {\in }_0} 2\pi \underset{0}{\overset{R}{\int }}\frac{rdr}{{\left(R^2+r^2\right)}^{\frac{3}{2}}}=\frac{q}{2{\in }_0} \left(1-\frac{1}{\sqrt{2}}\right)$

$\therefore$     Flux through curved surface $= -\frac{q}{2{\in }_0} \left(1-\frac{1}{\sqrt{2}}\right)$

Note: Flux through surface can be calculated using concept of solid angle.

$\mathrm{\Omega }=2\pi \left(1-\cos \theta \right)=2\pi \left(1-\frac{1}{\sqrt{2}}\right)$

$\therefore$    Solid angle subtended$=2\pi \left(1-\frac{1}{\sqrt{2}}\right)$

$\varphi$ for  $4\pi$ solid angle $=\frac{q}{{\in }_0}$

$\therefore \varphi$ for  $2\pi \left(1-\frac{1}{\sqrt{2}}\right)$ solid angle $=\frac{q}{4\pi {\in }_0} . 2\pi \left(1-\frac{1}{\sqrt{2}}\right)$

$=\frac{q}{2{\in }_0} \left(1-\frac{1}{\sqrt{2}}\right)$