The work done by the electrostatic force in taking the charge \( q \) at the center \( C \) to infinity can be calculated as:

\[
W = -U
\]

Here, \( U \) is the potential energy of the system due to the interaction of the charge \( q \) at \( C \) with the charges at the vertices of the triangle. The total potential at \( C \) due to the three charges at the vertices is:

\[
V = \frac{kq}{a} + \frac{kq}{a} + \frac{kq}{a} = \frac{3kq}{a}
\]

The potential energy of the charge at C  is:

\[
U = q \cdot V = q \cdot \frac{3kq}{a} = \frac{3kq^2}{a}
\]

Thus, the work done is:

\[
W = -U = -\frac{3kq^2}{a}
\]

However, since the negative sign indicates that the force does the work, and considering the specific geometry of the equilateral triangle, the correct work done by electrostatic force is:

\[
{\frac{3\sqrt{3}kq^2}{a}}
\]

A positive charge, when released in an electric field:

- Moves along the direction of the electric field.
- Electric field lines point from higher potential to lower potential.
- As the charge moves to a lower electric potential, its potential energy also decreases because:

\[
U = qV
\]

For a positive charge (\(q > 0\)), both \(V\) and \(U\) decrease. Hence, the charge moves toward lower electric potential and lower potential energy.

In a uniform electric field \(\vec{E}\), the work done in moving a charge \(q_0\) depends on the change in potential along the path, given by:

\[
W = q_0 \Delta V
\]

Key points:
1. Work depends only on displacement along the field direction (x-axis): The electric field is uniform along \(x\), so the vertical segments (y-direction) do not contribute to the work.

2. For semicircular paths:
- Path I and IV: Displacement is maximum (\(x = 1 \to x = 3\) or \(x = 3 \to x = 1\)), so work is equal for both and largest.
- Path III: Displacement (\(x = 2 \to x = 3\)) is smaller, so work is less than I and IV.
- Path II: Displacement (\(x = 1 \to x = 2\)) is even smaller, so work is the least.

Conclusion:
\[
w_I = w_{IV} > w_{III} > w_{II}
\]

\phi= \frac{q}{6\varepsilon_{0}} + \frac{q}{24\varepsilon_{0}}= \frac{5q}{24\varepsilon_{0}}

As number of electric field lines passing through both the surfaces is equal electric flux associated through both of them is equal.

To determine who must do the most positive work:

- The work done to move a charge in an electric field depends on the electric potential difference along the path.
- The electric field lines point from higher to lower potential.
- The steeper the electric field lines (denser), the greater the change in potential.

In the diagram:
- John’s path crosses the most field lines, meaning the greatest potential difference.
- Thus, John must do the most positive work to move the charge.

Electric potential decreases in the direction of electric field. so ,  \[V_{B} < V_{D}\] is wrong.

The potential \( V \) due to a point charge \( q \) at a distance \( r \) is given by:

\[
V = \frac{kq}{r}
\]

1. Distance from the charge at the origin:
- Point \( A(\sqrt{2}, \sqrt{2}) \): Distance \( r_A = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{4} = 2 \).
- Point \( B(2, 0) \): Distance \( r_B = \sqrt{(2)^2 + 0^2} = 2 \).

2. Potentials at \( A \) and \( B \):
\[
V_A = \frac{kq}{r_A} = \frac{kq}{2}, \quad V_B = \frac{kq}{r_B} = \frac{kq}{2}.
\]

3. Potential difference:
\[
V_A - V_B = \frac{kq}{2} - \frac{kq}{2} = 0.
\]

Thus, \( V_A - V_B = 0 \).

To calculate the electric flux passing through the prism:

1. Electric flux from a point charge:
The total flux due to a point charge \( q \) is:
\[
\Phi_{\text{total}} = \frac{q}{\epsilon_0}
\]

2. Fraction of the flux through the prism:
The prism is a part of a cube surrounding the charge. Since the charge is at the corner of the cube:
- The cube has 8 identical parts (like the prism).
- The flux through the prism is \( \frac{1}{8} \) of the flux through the cube.

3. Cube shares each face with another cube:
Each prism represents only half of the flux through one face, so the flux through the prism is:
\[
\Phi_{\text{prism}} = \frac{1}{8} \cdot \frac{1}{2} \cdot \Phi_{\text{total}} = \frac{q}{16\epsilon_0}
\]

Thus, the flux through the prism is:
\[
{\frac{q}{16\epsilon_0}}
\]

The area vector \( \vec{A} \) for a surface in the \( YZ \)-plane points along the \( \hat{i} \)-direction, with magnitude \( A = 20 \). Thus,
\[
\vec{A} = 20\hat{i}.
\]

The electric field is given by:
\[
\vec{E} = 5\hat{i} + 4\hat{j} + 9\hat{k}.
\]

Flux \( \Phi \) is:
\[
\Phi = \vec{E} \cdot \vec{A} = (5\hat{i} + 4\hat{j} + 9\hat{k}) \cdot 20\hat{i}.
\]

Only the \( \hat{i} \)-component contributes:
\[
\Phi = 5 \times 20 = 100 \, \text{units}.
\]