The flat base of a hemisphere of radius $$a$$ with no charge inside it lies in a horizontal plane. A uniform electric field $$\vec{E}$$ is applied at an angle $$\frac{\pi}{4}$$ with the vertical direction. The electric flux through the curved surface of the hemisphere is

Let the hemisphere of radius $$a$$ form a closed surface by adding its flat circular base. Since no charge is enclosed, Gauss’s law gives

$$\Phi_{\text{total}} = \Phi_{\text{curved}} + \Phi_{\text{base}} = 0$$

Hence

$$\Phi_{\text{curved}} = -\,\Phi_{\text{base}} \quad -(1)$$

The base is a circle of area $$A = \pi a^{2}$$ lying in the horizontal plane. Its outward normal is vertical and directed downwards (away from the interior of the hemisphere).

The uniform electric field $$\vec E$$ makes an angle $$\frac{\pi}{4}$$ with the vertical. Therefore, the angle between $$\vec E$$ and the outward normal to the base is also $$\frac{\pi}{4}$$.

Flux through the base:

$$\Phi_{\text{base}} = \vec E \cdot \vec A = EA\cos\left(\frac{\pi}{4}\right) = E\bigl(\pi a^{2}\bigr)\,\frac{1}{\sqrt{2}} = \frac{\pi a^{2}E}{\sqrt{2}}$$

Substituting this value in equation (1):

$$\Phi_{\text{curved}} = -\left(\frac{\pi a^{2}E}{\sqrt{2}}\right)$$

The minus sign only indicates that the flux through the curved surface is opposite in sense to that through the base. Its magnitude—the quantity usually asked for—is

$$\boxed{\dfrac{\pi a^{2}E}{\sqrt{2}}}$$

Thus the required electric flux through the curved surface of the hemisphere corresponds to Option B.