Electric Flux and Potential for Charge Outside Hemispherical Surface

To determine the correct statements, let us analyze each option individually using the principles of electrostatics and Gauss's Law.

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1. Analysis of Option [A] and Option [B] (Electric Flux)

Let us set up a coordinate system where the center of the flat circular base of the hemisphere is at the origin $O(0,0,0)$.

• The flat surface lies in the $xy$-plane ($z = 0$) with radius $R$, i.e., $x^2 + y^2 \le R^2, z = 0$.
• The curved surface is the upper hemisphere defined by $x^2 + y^2 + z^2 = R^2$ for $z \ge 0$.
• The point charge $+Q$ is located just outside the apex of the hemisphere at $(0,0,R^+)$.

Since the charge $+Q$ lies strictly outside the closed surface formed by the curved surface and the flat base, the total enclosed charge is $Q_{\text{encl}} = 0$.

By Gauss's Law: $\Phi_{\text{total}} = \Phi_{\text{curved}} + \Phi_{\text{flat}} = \frac{Q_{\text{encl}}}{\varepsilon_0} = 0$

Therefore, the total flux through the combined curved and flat surfaces is $0$. This shows that Statement [B] is incorrect.

To find the flux passing through the flat surface, consider the cone formed with vertex at $+Q(0,0,R)$ and base as the circular flat surface. The radius of this flat disk is $R$, and the distance from $+Q$ to the center of the disk is $R$.

The semi-vertical angle $\alpha$ of this cone is: $\tan \alpha = \frac{R}{R} = 1 \implies \alpha = \frac{\pi}{4}$

The solid angle $\Omega_{\text{flat}}$ subtended by the flat base at the position of $+Q$ is given by: $\Omega_{\text{flat}} = 2\pi (1 - \cos \alpha) = 2\pi \left(1 - \cos \frac{\pi}{4}\right) = 2\pi \left(1 - \frac{1}{\sqrt{2}}\right)$

The flux of the electric field originating from $+Q$ that passes through the flat surface is: $\Phi_{\text{flat}} = \frac{Q}{4\pi \varepsilon_0} \cdot \Omega_{\text{flat}} = \frac{Q}{4\pi \varepsilon_0} \cdot 2\pi \left(1 - \frac{1}{\sqrt{2}}\right) = \frac{Q}{2\varepsilon_0} \left(1 - \frac{1}{\sqrt{2}}\right)$

Since $\Phi_{\text{curved}} + \Phi_{\text{flat}} = 0$, the net electric flux passing through the curved surface is: $\Phi_{\text{curved}} = -\Phi_{\text{flat}} = -\frac{Q}{2\varepsilon_0} \left(1 - \frac{1}{\sqrt{2}}\right)$

Hence, Statement [A] is correct.

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2. Analysis of Option [C] (Normal Component of Electric Field)

Consider a point on the flat surface at a distance $r$ from the center $O(0,0,0)$, where $0 \le r \le R$.

The distance $d$ from the charge $+Q(0,0,R)$ to this point is: $d = \sqrt{r^2 + R^2}$

The magnitude of the electric field at this point is: $E = \frac{Q}{4\pi \varepsilon_0 d^2} = \frac{Q}{4\pi \varepsilon_0 (r^2 + R^2)}$

The normal component of the electric field (perpendicular to the flat surface, along the $z$-axis) is: $E_n = E \cos \theta = E \cdot \frac{R}{d} = \frac{Q R}{4\pi \varepsilon_0 (r^2 + R^2)^{3/2}}$

As $r$ varies from $0$ at the center to $R$ at the boundary, $E_n$ changes continuously. Thus, the normal component of the electric field is not constant over the flat surface.

Hence, Statement [C] is incorrect.

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3. Analysis of Option [D] (Equipotential Circumference)

The circumference of the flat base consists of all points lying on the circle $r = R, z = 0$.

The distance from $+Q(0,0,R)$ to any point on this circumference is: $d = \sqrt{R^2 + R^2} = \sqrt{2}R$

The electric potential $V$ at any point on the circumference due to the point charge $+Q$ is: $V = \frac{Q}{4\pi \varepsilon_0 (\sqrt{2}R)} = \text{constant}$

Since the potential is uniform at every point along the circumference, the circumference of the flat surface is indeed an equipotential line.

Hence, Statement [D] is correct.

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Conclusion

• Correct Options: A and D