A particle A having a charge of \( 2\times10^{-6}\,\mathrm{C} \) and a mass of 80 g is fixed at the bottom of a smooth inclined plane of inclination \( 30^\circ \). Where should another particle B, having the same charge and mass, be placed on the incline so that it may remain in equilibrium? (Take \( g=10\,\mathrm{m\,s^{-2}} \).)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 30 cm
The incline is smooth, so there is no friction. B stays at rest when the Coulomb repulsion of A, acting up the incline, balances the component of its weight along the incline, \( mg\sin30^\circ \).
In SI units the data are:
\( q=2\times10^{-6}\,\mathrm{C} \)
\( m=80\,\mathrm{g}=0.08\,\mathrm{kg} \)
\( k=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \)
The force needed along the incline is:
\( \begin{aligned} F &= mg\sin30^\circ \\ &= 0.08\,\mathrm{kg}\times10\,\mathrm{m\,s^{-2}}\times0.5 \\ &= 0.4\,\mathrm{N} \end{aligned} \)
\( \begin{aligned} kq^{2} &= 9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}\times(2\times10^{-6}\,\mathrm{C})^{2} \\ &= 0.036\,\mathrm{N\,m^{2}} \end{aligned} \)
Equating the Coulomb force at distance \( r \) to this force:
\( \displaystyle \frac{kq^{2}}{r^{2}}=mg\sin30^\circ \)
\( \begin{aligned} r^{2} &= \frac{kq^{2}}{mg\sin30^\circ} \\ &= \frac{0.036\,\mathrm{N\,m^{2}}}{0.4\,\mathrm{N}} \\ &= 0.09\,\mathrm{m^{2}} \end{aligned} \)
\( r=0.3\,\mathrm{m}=30\,\mathrm{cm} \)
Therefore, B must be placed 30 cm from A along the incline.