Explanation and Solution

The work done by an electrostatic field in moving a charge from one point to another depends only on the electric potential difference between the two points, since the electrostatic force is conservative.

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Given:
- We have a point charge \(q\) at the center of a circle.
- Points 1, 2, 3, and 4 lie on the same circle around the charge \(q\).

Step 1: Understand the electric potential
The electric potential \(V\) at any point on a circle centered around the charge \(q\) is the same because:

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

where \(r\) is the radius of the circle. Since points 1, 2, 3, and 4 are equidistant from \(q\), the potential at all these points is identical.

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Step 2: Work done in moving a charge
The work done \(W\) in moving a charge \(Q\) from one point to another in an electrostatic field is given by:

\[
W = Q (V_{\text{final}} - V_{\text{initial}})
\]

Since the potential \(V\) is the same at points 2, 3, and 4, the potential difference for each movement is zero:

\[
V_{\text{final}} = V_{\text{initial}}
\]

Thus, for movements from point 1 to points 2, 3, and 4, the work done:

\[
W_2 = W_3 = W_4 = 0
\]

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Final Comparison:
The work done \(W_2\), \(W_3\), and \(W_4\) are all equal. Hence:

\[
{W_2 = W_3 = W_4 = 0}
\]

This result arises because the electric field is conservative and the movement is along an equipotential surface.

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.

At the distance of closest approach \( r_0 \), the kinetic energy is completely converted into electrostatic potential energy: \( \frac{1}{2} m v^2 = \frac{1}{4 \pi \varepsilon_0} \frac{Ze^2}{r_0} \). Rearranging gives \( r_0 = \frac{Ze^2}{2\pi\varepsilon_0 m v^2} \).

Total charge is always conserved during sharing. Electrical potential energy may decrease and convert to heat, implying 'loss' of electrical energy (A is true). However, total energy is always conserved; (R) is false because energy conservation never fails.

For a free target, kinetic energy is shared, reducing the effective kinetic energy, leading to a larger closest approach distance (A is true). For a free target, mechanical energy of interacting particles is conserved. For a fixed target, external forces do work, so mechanical energy of the interacting particles is not conserved (R is true). This difference explains (A).

For a free target, the target can recoil, sharing kinetic energy with the projectile, leading to less relative kinetic energy converted into potential energy, hence a larger distance of closest approach. Total energy is always conserved if only conservative forces are present for both free and fixed targets. Thus (A) is true and (R) is false.

Assertion (A): Electrostatic potential energy between two positive charges is \(U = k \frac{q_1 q_2}{r}\). If they move away, \(r\) increases, thus \(U\) decreases. So (A) is true.

Reason (R): The change in potential energy is defined as the negative of the work done by conservative forces (like electrostatic force): \(Delta U = -W_{electrostatic}\). So (R) is false.nTherefore, (A) is true and (R) is false.