Question

A particle $A$ having a charge of $2\times {10}^{-6}C$ and a mass of $100\text{g}$ is fixed at the bottom of a smooth inclined plane of inclination ${30}^{\circ }$. Where should another particle $B$ having same charge and mass be placed from bottom on the inclined plane so that $B$ may remain in equilibrium ?

• A. $8$ $cm$
• B. $13$ $cm$
• C. $21$ $cm$
• D. $27$ $cm$

Answer 1

The correct option is D $27$ $cm$


From equilibrium of forces,
$F_e=mg\sin {30}^{\circ }$
$\Rightarrow \frac{k{q}^2}{r^2}=mg\sin {30}^{\circ }$
$\Rightarrow r=\sqrt{\frac{k{q}^2}{mg\sin {30}^{\circ }}}$
$\Rightarrow r=\sqrt{\frac{9\times {10}^9\times (2\times {10}^{-6}{)}^2}{100\times {10}^{-3}\times 10\times \frac{1}{2}}}$
$r=0.268\text{m}\approx 0.27\text{m}\approx 27\text{cm}$