Two identical conducting spheres with negligible volume have 2.1 nC and -0.1 nC charges, respectivel

Q: Why did the official answer key give 7.56 instead of 36?

The official answer key calculated the force *before* the spheres were brought into contact. This is a known error in the source material. Always trust the physics: if they touch, charges redistribute!

Q: What if the spheres were not identical?

If the spheres had different radii, the charge would distribute such that their surface potentials become equal ($V_1 = V_2$). The larger sphere would hold more charge, proportional to its radius.

Q: Why do we add the charges algebraically when they touch?

Charge is a fundamental conserved property of the universe. When the spheres touch, they form an isolated system, so the total charge before contact must exactly equal the total charge after contact.

Q: Does the volume of the spheres matter for Coulomb's law?

The question states they have 'negligible volume,' which allows us to treat them as perfect point charges. If they were large and close together, charge polarization would make Coulomb's law inaccurate.

Q: Why does the force change from attractive to repulsive?

Initially, the charges have opposite signs (+2.1 nC and -0.1 nC), so they attract. After contact, they share the net positive charge, both becoming +1.0 nC. Since like charges repel, the final force is repulsive.

Question

Two identical conducting spheres with negligible volume have $2.1 	ext{ nC}$ and $-0.1 	ext{ nC}$ charges, respectively. They are brought into contact and then separated by a distance of $0.5 	ext{ m}$. The electrostatic force acting between the spheres is ............... $	imes 10^{-9}$ N.
[Given, $4\pi\varepsilon_0 = \frac{1}{9 	imes 10^9}$ SI unit]

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine two tiny, identical conducting spheres. One carries a positive charge of $2.1 	ext{ nC}$, and the other carries a slight negative charge of $-0.1 	ext{ nC}$.

They are initially separated by a distance of $0.5 	ext{ m}$.

The problem asks for the electrostatic force between them *after* they are brought into contact and separated again. However, the official answer key provided a solution for the force *before* they touch. Let's explore both scenarios to build a rock-solid understanding!

### The Official Key's Perspective (Initial Force)

If we ignore the "brought into contact" part of the question, we simply apply **Coulomb's Law** to the initial charges.

The formula for the electrostatic force is:

$$F = \frac{1}{4\pi\varepsilon_0} \frac{|q_1 q_2|}{d^2}$$

We are given the constant $\frac{1}{4\pi\varepsilon_0} = 9 	imes 10^9 	ext{ N}\cdot	ext{m}^2/	ext{C}^2$.

Substituting our initial values into the equation:

$$F_{	ext{initial}} = \frac{9 	imes 10^9 	imes (2.1 	imes 10^{-9}) 	imes (0.1 	imes 10^{-9})}{(0.5)^2}$$

Calculating the numerator gives $1.89 	imes 10^{-9}$, and the denominator is $0.25$.

$$F_{	ext{initial}} = \frac{1.89 	imes 10^{-9}}{0.25} = 7.56 	imes 10^{-9} 	ext{ N}$$

This attractive force of **$7.56 	imes 10^{-9} 	ext{ N}$** matches the official answer key perfectly. But as elite physics students, we must follow the physical events described in the problem!

### The True Physics: Conduction and Charge Sharing

When two conducting spheres are brought into contact, they form a single equipotential body.

Because charge is a conserved quantity, the total net charge of the system remains constant. We add the initial charges algebraically:

$$Q_{	ext{net}} = 2.1 	ext{ nC} + (-0.1 	ext{ nC}) = 2.0 	ext{ nC}$$

Since the spheres are perfectly identical in size and shape, **symmetry dictates that they must share this net charge equally**.

Each sphere will take exactly half of the total charge:

$$q' = \frac{2.0 	ext{ nC}}{2} = 1.0 	ext{ nC}$$

Now, both spheres carry an identical positive charge of $1.0 	ext{ nC}$.

### The Final Calculation

The spheres are now separated back to their original distance of $0.5 	ext{ m}$.

We apply Coulomb's Law one more time using our new, redistributed charges:

$$F_{	ext{final}} = \frac{9 	imes 10^9 	imes (1.0 	imes 10^{-9}) 	imes (1.0 	imes 10^{-9})}{(0.5)^2}$$

$$F_{	ext{final}} = \frac{9 	imes 10^{-9}}{0.25} = 36 	imes 10^{-9} 	ext{ N}$$

Because both spheres now carry positive charges, this true final force is **repulsive**.

While the official key expected the initial force calculation, understanding the profound principle of **charge conservation and equipotential sharing** is what truly prepares you for the toughest JEE challenges!

Enter Numerical Value:

Visualized Solution

Initial Setup

Coulomb's Law

Initial Force Calculation

Initial Force Result

Spheres in Contact

Charge Redistribution

Final Force Setup

True Final Force

The Sigma Insight: Coulomb's Law

Analyzing the Setup

The Official Key's Perspective (Initial Force)

The True Physics

The Final Calculation

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