Two small spheres A and B carry charges \(+2\ \mu\mathrm{C}\) and \(+3\ \mu\mathrm{C}\). They are held 30 cm apart in vacuum. Sphere B has a mass of 2 g. Ignore gravity. Sphere B is now released. Its acceleration just after release is (take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(300\ \mathrm{m\,s^{-2}}\)
Sphere B feels the Coulomb force of sphere A. Just after release, this force gives B an acceleration \( a=F/m \).
Given:
\( q_1=2\times10^{-6}\ \mathrm{C} \)
\( q_2=3\times10^{-6}\ \mathrm{C} \)
\( r=30\ \mathrm{cm}=0.30\ \mathrm{m} \)
\( m=2\ \mathrm{g}=2\times10^{-3}\ \mathrm{kg} \)
\( \begin{aligned} F &= \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^{2}} \\ &= \frac{9\times10^{9}\times2\times10^{-6}\times3\times10^{-6}}{(0.30)^{2}} \\ &= \frac{5.4\times10^{-2}}{9\times10^{-2}} \\ &= 0.6\ \mathrm{N} \end{aligned} \)
\( \begin{aligned} a &= \frac{F}{m} \\ &= \frac{0.6}{2\times10^{-3}} \\ &= 300\ \mathrm{m\,s^{-2}} \end{aligned} \)
Therefore, sphere B starts to move away from A with an acceleration of \( 300\ \mathrm{m\,s^{-2}} \), since like charges repel.
NCERT (2026-27):
Coulomb’s Law – “… Coulomb’s law is written as”
\( \displaystyle F=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^{2}} \)\( \qquad (1.2) \)
Example 1.3 – “Using Newton’s second law of motion, F = ma, the acceleration that an electron will undergo is …”