Q.A proton is kept at rest. A positively charged particle is released from rest at a distance $d$ in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time $t$, the work done on the two moving charged particles is

Concept: Work-Energy Theorem and Electrostatic Repulsion

Both particles experience the same repulsive Coulomb force initially ($F = k\frac{e^2}{d^2}$), but their accelerations differ because mass enters Newton's second law: $a = F/m$. The positron, being ~1836 times lighter than the proton, accelerates far more rapidly.

In time $t$, the positron travels a much larger distance than the proton. Since both start from the same point and experience repulsive forces that decrease with distance, the positron quickly moves into regions where the force is weaker. However, work done is $W = \int F \, dx$. The positron's larger displacement more than compensates for the weakening force over most of its trajectory.

By the work-energy theorem, $W = \Delta KE = \frac{1}{2}mv^2$. The positron's smaller mass means that for the same kinetic energy, it needs a much higher velocity—which it achieves by traveling farther under the (on average, similar) force field. The integral $\int F \, dx$ is larger when $x$ is larger, even though $F$ falls off.

The heart of this problem lies in understanding that work is not force alone—it is $W = \int \vec{F} \cdot d\vec{s}$, the accumulation of force over the actual path traveled. Two particles experiencing the same repulsive Coulomb force will do different amounts of work if their masses differ, because the lighter particle accelerates more and covers more ground.

Both the proton and the positron carry charge $+e$ and are released from rest at distance $d$ from a stationary proton. The electrostatic repulsion obeys Coulomb's law:

where $r$ is the instantaneous separation. At any given position $r$, both particles feel identical force. But their responses—their accelerations—are wildly different.

Step-by-step reasoning

1. Acceleration depends on mass.

   Newton's second law gives $a = F/m$. The proton has mass $m_p \approx 1836 \, m_e$, while the positron has mass $m_e$. At the same distance, the positron's acceleration is roughly 1836 times larger than the proton's.

2. Greater acceleration means greater displacement in the same time.

   Starting from rest, a particle with larger acceleration will travel farther in time $t$. The positron races away from the stationary proton much faster than a second proton would.

3. Work is force integrated over displacement.

Even though $F(r)$ is the same function for both particles, the upper limit $r(t)$—the position reached at time $t$—is much larger for the positron. The positron samples the repulsive force over a longer stretch of its journey.

4. The force weakens with distance, but not fast enough to reverse the trend.

   Yes, as the positron moves farther out, $F \propto 1/r^2$ drops. But the integral of force over distance—which is work—still grows because you are adding more (albeit smaller) contributions. The positron's head start in displacement more than compensates for the weakening force.

   Alternatively, invoke energy: the work done equals the kinetic energy gained, $W = \frac{1}{2}m v^2$. For the same force profile, the lighter positron reaches a higher speed and hence higher kinetic energy in time $t$.

5. Why the other options fail:

   • (A) claims the work is the same because the force law is the same. But work depends on displacement, not just the form of $F(r)$.
   • (B) suggests the positron does less work because the force weakens. This confuses instantaneous force with integrated work; the larger displacement dominates.
   • (D) invokes Newton's third law: the moving particle exerts an equal and opposite force on the stationary proton. However, the stationary proton does not move (it is held at rest), so zero displacement means zero work done on it. Work is not the same.

Concept: Work Depends on Force AND Displacement, Not Force Alone

Step 1: Note the force is identical in both experiments

Both the proton and the positron carry charge $+e$, so $F(r) = ke^2/r^2$ is the same function of separation $r$ in both cases.

Step 2: Compare accelerations via Newton's second law

The positron's mass is $\approx 1/1836$ of the proton's, so at any given separation its acceleration is $\approx 1836\times$ larger.

Step 3: Compare displacement in the same time $t$

Starting from rest, the much larger acceleration means the positron covers a much larger distance than the proton does in the same time $t$.

Step 4: Compare work via $W=\int F\,dr$

Even though $F(r)$ weakens with $r$, the positron sweeps through the (weakening) force field over a much longer path in the same time — the extra distance dominates over the weakening force, so more energy (equivalently $\Delta K$) is transferred to the positron.

Final Answer:

Option (c): more for the positron, because it moves away a larger distance in the same time