\(\phi=\oint E d S=E \oint d S \cos 45^{\circ}\)
In case of hemisphere
\(\phi_{\text {curved }}=\dot{\phi}_{\text {cirrular }}\)
Therefore. \(\phi_{\text {curved }}=E \pi a^{2} \cdot \frac{1}{\sqrt{2}}=\frac{E \pi a^{2}}{\sqrt{2}}\)
(Take $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}$ )