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.

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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:

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

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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:

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

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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:

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

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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 given problem involves calculating the work done by an external agent in moving a charge \(-Q\) from point \(A\) to the center \(O\) of a charged ring.

Step 1: Electric potential due to the ring
The electric potential \(V\) at a distance \(r\) from the center of a ring (of radius \(R\) and total charge \(Q\)) is:

\[
V(r) = \frac{kQ}{\sqrt{R^2 + r^2}}
\]

- At point \(A\) (\(r = \sqrt{35}R\)):

\[
V_A = \frac{kQ}{\sqrt{R^2 + (\sqrt{35}R)^2}} = \frac{kQ}{\sqrt{R^2 + 35R^2}} = \frac{kQ}{6R}
\]

- At the center \(O\) (\(r = 0\)):

\[
V_O = \frac{kQ}{R}
\]

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Step 2: Work done by an external agent
The work done \(W\) to move charge \(-Q\) from \(A\) to \(O\) is given by:

\[
W = -q(V_O - V_A)
\]

Substituting \(q = -Q\), \(V_O = \frac{kQ}{R}\), and \(V_A = \frac{kQ}{6R}\):

\[
W = -(-Q) \left(\frac{kQ}{R} - \frac{kQ}{6R}\right)
\]

\[
W = Q \left(\frac{kQ}{R} - \frac{kQ}{6R}\right)
\]

\[
W = Q \cdot \frac{6kQ - kQ}{6R} = Q \cdot \frac{5kQ}{6R}
\]

\[
W = \frac{-5kQ^2}{6R}
\]

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Final Answer:
\[
{\frac{-5kQ^2}{6R}}
\]

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:

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

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

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.

This statement applies when a particle moves against a force field (e.g., an electron moving against an electric field). As it moves, work is done on it, leading to:

- Decrease in kinetic energy: The particle slows down as it loses speed.
- Increase in potential energy: The work done on the particle increases its potential energy.

This satisfies the conservation of energy principle.

The kinetic energy gained by the electron is equal to the work done by the electric field:

\[
eV = \frac{1}{2} m v^2
\]

Rearranging for \(v\):

\[
v = \sqrt{\frac{2eV}{m}}
\]

This shows that \(v \propto \sqrt{V}\).