Question

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 N. [Given : $4\pi\epsilon_0 = \frac{1}{9 \times 10^9}$ SI unit]

Hint:

For identical conducting spheres, when they touch, the final charge on each is simply the average of their initial charges: $q_{final} = (q_1 + q_2)/2$. This is a direct application of conservation of charge and potential equalization.

Answer 1

Correct Answer: 36

To solve this problem, we first need to determine the charge on each sphere after they are brought into contact. Once they touch, the charge will redistribute equally because the spheres are identical. We first find the total initial charge:

• qinitial=2.1+−0.1=2.0nC

After redistribution, each sphere has: 

• qfinal=2.02=1.0nC

Next, we calculate the electrostatic force using Coulomb's law:

F=kq·qr2

where q=1.0 10-9C, r=0.5m, and k=9 10⁹

Substitute these values into the formula:

F=9 10⁹  (1.0 10^-9)²0.52

Calculate 1.0-9  C=1   nC:

• The force becomes:

F=9 10⁹=1     0.52

=9 10⁹4e2

Solving gives:

F^-9=36 10^-9N

This force is clearly within the given range of 36,36. Therefore, the electrostatic force is indeed 36 NN.