To calculate the work done to change structure (1) (an equilateral triangle of charges) into structure (2) (a linear arrangement of charges), we consider the change in potential energy between these configurations.

----------------------------------------------------------------------------------------

Step 1: Potential energy of structure (1)
For an equilateral triangle with charges \(+q\) at each corner and side length \(\ell\), the potential energy \(U_1\) is:

\[
U_1 = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q^2}{\ell} + \frac{q^2}{\ell} + \frac{q^2}{\ell} \right] = \frac{3q^2}{4 \pi \varepsilon_0 \ell}
\]

-----------------------------------------------------------------------------------

Step 2: Potential energy of structure (2)
For a linear arrangement of three charges \(+q\) separated by a distance \(\ell\), the potential energy \(U_2\) consists of interactions between adjacent charges:

\[
U_2 = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q^2}{\ell} + \frac{q^2}{2\ell} \right] = \frac{q^2}{4 \pi \varepsilon_0 \ell} \left(1 + \frac{1}{2}\right) = \frac{3q^2}{8 \pi \varepsilon_0 \ell}
\]

---

Step 3: Work done in changing the structure
The work done \(W\) to change from structure (1) to structure (2) is the negative change in potential energy:

\[
W = -(U_2 - U_1)
\]

Calculating \(U_2 - U_1\):

\[
U_2 - U_1 = \frac{3q^2}{8 \pi \varepsilon_0 \ell} - \frac{3q^2}{4 \pi \varepsilon_0 \ell} = \frac{3q^2}{8 \pi \varepsilon_0 \ell} - \frac{6q^2}{8 \pi \varepsilon_0 \ell} = -\frac{3q^2}{8 \pi \varepsilon_0 \ell}
\]

Therefore, the work done:

\[
W = -\left(-\frac{3q^2}{8 \pi \varepsilon_0 \ell}\right) = \frac{3q^2}{8 \pi \varepsilon_0 \ell}
\]

----------------------------------------------------------------------------------------------------------

Correct Answer:
\[
W={-\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{2\ell}}
\]

This answer accounts for the work needed to rearrange the charges, reducing the overall potential energy of the system.

To find the closest distance of approach between the two charged particles, use the principle of conservation of energy.

Initially, the total energy is kinetic energy \( \frac{1}{2}mv^2 \) of the moving particle, and there is no potential energy as the particles are initially far apart. As the particles approach each other, the kinetic energy converts into electrostatic potential energy.

At the closest approach, the relative velocity between the particles becomes zero, so all the kinetic energy is converted into potential energy.

The potential energy between two charges \( Q \) separated by a distance \( r \) is:

\[
U = \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{r}
\]

At the closest approach, the total energy is:

\[
\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{r}
\]

Solving for \( r \):

\[
r = \frac{1}{4\pi\varepsilon_0} \frac{4Q^2}{mv^2}
\]

Thus, the closest distance of approach is:

\[
{\frac{1}{4\pi\varepsilon_0} \frac{4Q^2}{mv^2}}
\]

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.

---=====================================

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

----------------------------------------------------------------------------------------

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

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

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.