In free space, a particle A of charge 1 micro C is held fixed at a point P. Another particle B of th

Q: Why doesn't particle A move when they repel each other?

The problem explicitly states that particle A is "held fixed" at point P. This means an external force is keeping it in place, so only particle B is free to accelerate and gain kinetic energy.

Q: Why do we use conservation of energy instead of calculating the force and acceleration?

The electrostatic force $F = \frac{K q_1 q_2}{r^2}$ changes continuously as the distance $r$ increases. This means the acceleration is not constant, making kinematic equations ($v^2 = u^2 + 2as$) invalid. Energy conservation elegantly bypasses this varying force.

Q: I got $v^2 = 4 \times 10^9$. Why is the answer $2 \times 10^3$?

You did the math perfectly! The original JEE question contains a typo. The mass was given as $4\mu\text{g}$ ($10^{-9}\text{ kg}$), which leads to $v^2 = 4 \times 10^9$. However, the options were designed assuming a mass of $4\text{ mg}$ ($10^{-6}\text{ kg}$), which gives $v^2 = 4 \times 10^6$ and $v = 2 \times 10^3\text{ m/s}$.

Q: Does the potential energy ever become zero?

The potential energy $U = \frac{K q_1 q_2}{r}$ approaches zero only as the distance $r$ approaches infinity. At any finite distance, there is still some potential energy left in the system.

Q: What would happen if particle B had a negative charge?

If particle B were negative, the force would be attractive. If released from rest, it would move *towards* particle A, not away from it. The potential energy would become more negative, and its kinetic energy would increase as it gets closer.

Question

In free space, a particle $A$ of charge $1\mu	ext{C}$ is held fixed at a point $P$. Another particle $B$ of the same charge and mass $4\mu	ext{g}$ is kept at a distance of $1	ext{ mm}$ from $P$. If $B$ is released, then its velocity at a distance of $9	ext{ mm}$ from $P$ is
$\left[	ext{Take}, \frac{1}{4\pi\varepsilon_0} = 9 	imes 10^9 	ext{ N-m}^2	ext{C}^{-2}ight]$

Options:

(A) $1.5 	imes 10^2 	ext{ m/s}$

(B) $3.0 	imes 10^4 	ext{ m/s}$

(C) $1.0 	ext{ m/s}$

(D) $2.0 	imes 10^3 	ext{ m/s}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are in the vast emptiness of free space. There is no gravity, no air resistance, just pure, unadulterated physics. We pin down a particle, let's call it $A$, at a fixed point $P$. This particle carries a positive charge of $1\mu	ext{C}$.

Now, we bring in a second particle, $B$, which also carries a positive charge of $1\mu	ext{C}$. We hold it exactly $1	ext{ mm}$ away from $A$. Because both particles are positively charged, they despise each other. The electrostatic force of repulsion is screaming at them to push apart.

### The Master Equation

The moment we release particle $B$, it accelerates away from $A$. As it moves, the distance between them increases, and the electrostatic potential energy of the system decreases. But energy cannot simply vanish! The universe demands balance.

This lost potential energy is entirely converted into the kinetic energy of particle $B$. This is the beautiful principle of **Conservation of Mechanical Energy**. Mathematically, we write this as:
$$U_i + K_i = U_f + K_f$$

Since particle $B$ starts from rest, its initial kinetic energy $K_i$ is zero. Thus, the gain in kinetic energy is simply the difference in potential energy:
$$\Delta K = U_i - U_f$$

### Calculating the Energies

Let's calculate the initial potential energy, $U_1$, when the separation is $r_1 = 1	ext{ mm}$. The formula for electrostatic potential energy is:
$$U = \frac{K q_A q_B}{r}$$

Substituting our known values:
$$U_1 = \frac{(9 	imes 10^9) (1 	imes 10^{-6}) (1 	imes 10^{-6})}{1 	imes 10^{-3}}$$

Notice how the powers of ten interact. In the numerator, $10^9 	imes 10^{-6} 	imes 10^{-6} = 10^{-3}$. 
$$U_1 = \frac{9 	imes 10^{-3}}{10^{-3}} = 9	ext{ J}$$
The system initially holds a massive $9	ext{ J}$ of stored energy!

Now, particle $B$ flies outward and reaches a distance of $r_2 = 9	ext{ mm}$. Let's find the final potential energy, $U_2$, at this new position.
$$U_2 = \frac{(9 	imes 10^9) (1 	imes 10^{-6}) (1 	imes 10^{-6})}{9 	imes 10^{-3}}$$
$$U_2 = \frac{9 	imes 10^{-3}}{9 	imes 10^{-3}} = 1	ext{ J}$$

### The Kinetic Energy and The Typo

The potential energy has dropped from $9	ext{ J}$ to $1	ext{ J}$. That means exactly $8	ext{ J}$ of energy has been converted into kinetic energy.
$$\frac{1}{2} m_B v_B^2 = 8	ext{ J}$$

Now, we must substitute the mass of particle $B$. The problem states the mass is $4\mu	ext{g}$. In standard SI units, $4\mu	ext{g} = 4 	imes 10^{-9}	ext{ kg}$. Let's plug this in:
$$\frac{1}{2} (4 	imes 10^{-9}) v_B^2 = 8$$
$$2 	imes 10^{-9} v_B^2 = 8$$
$$v_B^2 = 4 	imes 10^9$$

Here is where we hit a fascinating roadblock. If $v_B^2 = 4 	imes 10^9$, then $v_B = \sqrt{40 	imes 10^8} \approx 6.32 	imes 10^4	ext{ m/s}$. But look at the options! None of them match this value.

**What happened?** 
This is a classic typo in the original JEE paper. The mass was intended to be $4	ext{ mg}$ (milligrams), not $4\mu	ext{g}$ (micrograms). Let's see what happens if we use $m_B = 4	ext{ mg} = 4 	imes 10^{-6}	ext{ kg}$:
$$\frac{1}{2} (4 	imes 10^{-6}) v_B^2 = 8$$
$$2 	imes 10^{-6} v_B^2 = 8$$
$$v_B^2 = 4 	imes 10^6$$

Taking the square root now gives a perfect, clean integer:
$$v_B = 2 	imes 10^3	ext{ m/s}$$

This perfectly matches option (d). It is crucial to trust your math. When you encounter a situation like this in an exam, quickly check if a common unit typo (like milli vs micro) leads to one of the options.

### Final Conclusion

By understanding the flow of energy and keeping a sharp eye on our units, we successfully navigated both the physics of the problem and the hidden trap set by the examiners. The final velocity of particle $B$ is $2.0 	imes 10^3	ext{ m/s}$.

Select Answer:

Visualized Solution

Initial Setup

Energy Conservation

Initial Potential Energy $U_{1}$

Calculating $U_{1}$

Final Potential Energy $U_{2}$

Calculating $U_{2}$

Change in Energy $Δ U$

Mass Conversion & Typo

Solving for $v_{B}$

What if $q_{A} < 0$ ?

The Sigma Insight: Electrostatic Potential Energy

Analyzing the Setup

The Master Equation

Calculating the Energies

The Kinetic Energy and The Typo

Final Conclusion

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