Read the following passage and answer the questions.
Coulomb used a torsion balance to study the force between two small charged metal spheres. A torsion balance is a very sensitive device for measuring small forces. The spheres were small compared with the distance between them, so they could be treated as point charges. At the start, Coulomb did not know the charge on either sphere.
To change a charge in a known way, he touched a charged sphere with an identical uncharged sphere. Then he took the second sphere away. He repeated this step to get smaller and smaller charges. In one set of readings, he kept the charges fixed and changed the distance between the spheres. In another set, he kept the distance fixed and changed the charges in pairs. By comparing all the readings, he arrived at a simple law for the force between two point charges.
In one reading, the two spheres carry equal charges and are 30 cm apart. The force between them is \(4\times10^{-5}\ \mathrm{N}\). What is the charge on each sphere? (Take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2\times10^{-8}\ \mathrm{C}\)
For two equal point charges \( q \) at a distance \( r \), Coulomb’s law gives \( F=kq^{2}/r^{2} \), with \( k=1/4\pi\varepsilon_0 \).
Given:
\( F=4\times10^{-5}\ \mathrm{N} \)
\( r=30\ \mathrm{cm}=0.3\ \mathrm{m} \)
\( \begin{aligned} q^{2} &= \frac{Fr^{2}}{k} \\ &= \frac{4\times10^{-5}\times0.09}{9\times10^{9}} \\ &= 4\times10^{-16}\ \mathrm{C^{2}} \end{aligned} \)
\( \begin{aligned} q &= \sqrt{4\times10^{-16}} \\ &= 2\times10^{-8}\ \mathrm{C} \end{aligned} \)
Therefore, each sphere carries a charge of \( 2\times10^{-8}\ \mathrm{C} \).
NCERT (2026-27):
Coulomb’s Law – “… Thus, if two point charges q1, q2 are separated by a distance r in vacuum, the magnitude of the force (F) between them is given by”
\( \displaystyle F=k\frac{|q_1q_2|}{r^{2}} \)\( \qquad (1.1) \)