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(D) The solid angle subtended by the flat circular base at the position of the charge $+Q$ is given by $\Omega = 2\pi(1 - \cos	heta)$.In the given ge

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$A, D$

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(D) The solid angle subtended by the flat circular base at the position of the charge $+Q$ is given by $\Omega = 2\pi(1 - \cos	heta)$.In the given geometry,the charge is at the pole of the hemisphere,so the angle $	heta$ subtended by the radius of the base at the charge is $45^{\circ}$.Thus,$\Omega = 2\pi(1 - \cos 45^{\circ}) = 2\pi(1 - \frac{1}{\sqrt{2}})$.The flux through the flat surface is $\Phi_{flat} = \frac{Q}{\varepsilon_0} 	imes \frac{\Omega}{4\pi} = \frac{Q}{2\varepsilon_0}(1 - \frac{1}{\sqrt{2}})$.Since the charge is outside the closed hemisphere,the net flux through the entire surface is zero. Therefore,the flux through the curved surface is $\Phi_{curved} = -\Phi_{flat} = -\frac{Q}{2\varepsilon_0}(1 - \frac{1}{\sqrt{2}})$. Statement $A$ is correct.Statement $B$ is incorrect because the total flux through a closed surface not enclosing the charge is zero.Statement $C$ is incorrect because the electric field varies with distance from the charge.Statement $D$ is correct because all points on the circumference of the flat base are at the same distance $R$ from the point charge $+Q$,making the potential $V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R}$ constant along the circumference.

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