Two identical conducting spheres with negligible volume have 2.1 nC and -0.1 nC charges, respectively.  They are brought into contact and then separated by a distance of 0.5 m.  The electrostatic force acting between the spheres is ___ $\times {10}^{-9}\mathrm{N}$

$\text{ [Given : }4{\mathrm{\pi \epsilon }}_0=\frac{1}{9\times {10}^9}\text{ SI unit] }$

The question involves calculating the electrostatic force acting between two identical conducting spheres after they are brought into contact and then separated by a distance.

Understanding Coulomb's Law

Coulomb's law states that the force of attraction or repulsion between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:

F = (K × q₁ × q₂) / r²

Where:

• q₁ = charge of the first sphere
• q₂ = charge of the second sphere
• r = distance between the two identical spheres
• K = Coulomb's constant = 9 × 10⁹ Nm²/C²

Calculation Steps

Given:

• Charge of the first sphere: q₁ = 2.1 nC
• Charge of the second sphere: q₂ = -0.1 nC
• Distance between the two identical spheres: r = 0.5 m

When the two identical conducting spheres are brought into contact, they share their charges equally. The total charge is distributed between the spheres, and the charge on each sphere becomes:

q = (q₁ + q₂) / 2

Substituting the values:

q = (2.1 nC + (-0.1 nC)) / 2

q = 2.0 nC / 2

q = 1 nC

Applying Coulomb's Law

After the charges are redistributed, the force between the two identical spheres is calculated as:

F = (K × q × q) / r²

Substitute the known values:

F = (9 × 10⁹ × 1 × 10^-9 × 1 × 10^-9) / (0.5)²

F = (9 × 10⁹ × 10^-18) / 0.25

F = 36 × 10^-9 N

Final Answer

Hence, the electrostatic force acting between the two identical conducting spheres is:

F = 36 × 10^-9 N