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 ______ × 10^-9 N. 

[Given \(4\pi {\varepsilon _0} = \frac{1}{{9 \times {{10}^9}}}\) SI unit] 

CONCEPT:

According to Coulomb's law, the force of attraction or repulsion is directly proportional to the product of distance between them and inversely proportional to the square of the distance between them.

F = K\(\frac{q_1q_2}{r^2}\)

Here we have, q1 is the charge of the first mass, q2 is the charge of the second mass, r is the distance and K is the proportionally constant.

CALCULATION:

Given: Charge of first sphere q₁ =  2.1 nC 

and Charge of second sphere q₂ = - 0.1 NC 

when they connect with each other the charge becomes, q = \(\frac{q_2-q_1}{2}\)

⇒  q = \(\frac{2.1-0.1}{2}\)

⇒ q = 1 nC

According to Coulomb's law, we have;

F = \(\frac{1}{4 \pi \epsilon_o}{\frac{q_1q_2}{r^2}}\)

Now, on putting the given values we have;

F = \(9 × 10^9 × {\frac{1 × 10^{-9}× 1 × 10^{-9}}{0.5^2}}\)

F = \(9 × 10^9 × {\frac{ 10^{-18}}{0.25}}\)

⇒ F = 36 \(× 10^{-9}\) N

Hence, F = 36 × 10^-9 N