The electric flux through a spherical surface depends only on the total enclosed charge, as per Gauss's law:

\[
\Phi = \frac{Q_{\text{enc}}}{\epsilon_0}
\]

Since the total charge enclosed remains the same regardless of the sphere's radius, the flux through the concentric sphere of radius 20 cm will also be 25 V-m.

Angle between area vector and Electric field is 90º. So Flux is zero.

To calculate the electric flux through the lateral surface of the vessel, we apply Gauss's law and consider the symmetry of the system.

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Step-by-Step Solution:

1. Total Flux from the Charge \(q\):
The total flux emitted by the point charge \(q\) in all directions is:
\[
\Phi_{\text{total}} = \frac{q}{\varepsilon_0}.
\]

2. Flux Through the Circular Mouth:
Using the geometry of the vessel, the charge \(q\) is placed at a depth \(h = \sqrt{3}R\). The flux through the circular mouth of radius \(R\) can be calculated as:
\[
\Phi_{\text{mouth}} = \frac{q}{2\varepsilon_0}.
\]

3. Flux Through the Lateral Surface
By symmetry, the flux through the lateral surface is the remaining flux from the total flux after subtracting the flux through the mouth:
\[
\Phi_{\text{lateral}} = \Phi_{\text{total}} - \Phi_{\text{mouth}}.
\]

4. Simplify:
Substitute the values:
\[
\Phi_{\text{lateral}} = \frac{q}{\varepsilon_0} - \frac{q}{2\varepsilon_0}.
\]
\[
\Phi_{\text{lateral}} = \frac{q}{2\varepsilon_0} \left(1 + \frac{\sqrt{3}}{2}\right).
\]

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Final Answer:
The electric flux through the lateral surface of the vessel is:
\[
{\frac{q}{2\varepsilon_0} \left( 1 + \frac{\sqrt{3}}{2} \right)}.
\]

We are tasked with finding the charge enclosed within the cube using Gauss's Law:

\[
\Phi_{\text{net}} = \frac{q_{\text{enc}}}{\varepsilon_0}.
\]

Given:
- Net flux through the cube: \(\Phi_{\text{net}} = 1.05 \, \text{Nm}^2\text{C}^{-1}\),
- \(\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2\text{N}^{-1}\text{m}^{-2}\),
- \(\alpha = 800 \, \text{N}\text{C}^{-1}\text{m}^{-1/2}\),
- Cube side length: \(a = 0.1 \, \text{m}\).

Steps:

1. Use Gauss's Law:
The enclosed charge \(q_{\text{enc}}\) is related to the flux by:
\[
q_{\text{enc}} = \varepsilon_0 \Phi_{\text{net}}.
\]

2. Substitute the values:
\[
q_{\text{enc}} = (8.85 \times 10^{-12}) \cdot (1.05).
\]

3. Calculate:
\[
q_{\text{enc}} = 9.27 \times 10^{-12} \, \text{C}.
\]

Final Answer:
The enclosed charge is:
\[
{9.27 \times 10^{-12} \, \text{C}}.
\]

Net Electric Point at Point C will be directed downwards

\[ \overrightarrow{E}=\frac{\sigma_{A}}{2\varepsilon_{0}} + \frac{\sigma_{B}}{2\varepsilon_{0}}\]

To calculate the flux \(\Phi_E\) of the electric field \(\vec{E}\) through the square, we use the formula:

\[
\Phi_E = \vec{E} \cdot \vec{A}
\]

Here:
- \(\vec{E} = 2 \times 10^3 \, \text{N/C}\) (along the positive x-axis),
- \(\vec{A}\) is the area vector, perpendicular to the plane of the square.

Since the square lies in the yz-plane, its area vector points along the x-axis (same direction as \(\vec{E}\)), and its magnitude is the area of the square:

\[
\text{Side of square} = 10 \, \text{cm} = 0.1 \, \text{m}, \quad \text{Area} = (0.1)^2 = 0.01 \, \text{m}^2
\]

The flux is:

\[
\Phi_E = |\vec{E}| \cdot |\vec{A}| \cdot \cos\theta
\]

Here, \(\theta = 0^\circ\) (since \(\vec{E}\) is parallel to \(\vec{A}\)):

\[
\Phi_E = (2 \times 10^3) \cdot 0.01 \cdot \cos 0^\circ = 20 \, \text{N·m}^2/\text{C}
\]

Thus, the flux is:

\[
{20 \, \text{N·m}^2/\text{C}}
\]

Electric field at Point II will be due to both the sheets so,

\[\overrightarrow{E}=\frac{\sigma}{2\varepsilon_{0}} + \frac{\sigma}{2\varepsilon_{0}}=\frac{\sigma}{\varepsilon_{0}}\]

To calculate the electric flux through the square, we use Gauss's law and symmetry principles.

The total flux from a point charge \(q = +20 \, \mu\text{C}\) is:

\[
\Phi_{\text{total}} = \frac{q}{\epsilon_0}
\]

Here:
- \(q = 20 \times 10^{-6} \, \text{C}\),
- \(\epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N·m}^2\).

The square is part of an imaginary cube with the charge at its center. The flux through one face of the cube (the square in question) is:

\[
\Phi_{\text{square}} = \frac{\Phi_{\text{total}}}{6} = \frac{q}{6\epsilon_0}
\]

Substitute the values:

\[
\Phi_{\text{square}} = \frac{20 \times 10^{-6}}{6 \times 8.854 \times 10^{-12}}
\]

Simplify:

\[
\Phi_{\text{square}} = 3.77 \times 10^5 \, \text{N·m}^2/\text{C}
\]

Thus, the flux through the square is approximately:

\[
{3.8 \times 10^5 \, \text{N·m}^2/\text{C}}
\]

The  torque acting on the dipole in an electric field is given by:

\[
\tau = pE \sin\theta
\]

The potential energy of the dipole is defined as:

\[
U = -pE \cos\theta
\]

Here, the potential energy is zero when \( \theta = 90^\circ \), which aligns with the condition provided.

Given:
- Dipole moment: \( \overrightarrow{p} = 3\hat{i} + 4\hat{j} \) C·m
- Electric field: \( \overrightarrow{E} = 0.4 \, \text{kN/C} \hat{i} = 400 \, \text{N/C} \hat{i} \)

Torque (\( \overrightarrow{\tau} \)):
\[
\overrightarrow{\tau} = \overrightarrow{p} \times \overrightarrow{E}
\]

\[
\overrightarrow{\tau} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
3 & 4 & 0 \\
400 & 0 & 0
\end{vmatrix} = \hat{k} \big(3(0) - 4(400)\big) = -1600\hat{k} \, \text{N·m}
\]

Potential Energy (\( U \)):
\[
U = -\overrightarrow{p} \cdot \overrightarrow{E}
\]
\[
U = -(3 \times 400 + 4 \times 0) = -1200 \, \text{J}
\]

Final Answer:
- Torque: \( -1600 \, \text{N·m} \hat{k} \)
- Potential energy: \( -1200 \, \text{J} \)