Question

A particle $A$ having a charge of $2.0\times {10}^{-6}\text{C}$ 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 on the inclined plane so that $B$ may remain in equilibrium ?

• A. $8\text{cm}$ from the bottom
• B. $13\text{cm}$ from the bottom
• C. $21\text{cm}$ from the bottom
• D. $27\text{cm}$ from the bottom

Answer 1

The correct option is D $27\text{cm}$ from the bottom

Given that,
Charge of the particle, $q=2.0\times {10}^{-6}\text{C}$
Mass of the particle, $m=100\text{g}=0.1\text{kg}$
The free body diagram of the system is given below

Charges $A$ and $B$ have same mass and same charge, thus electrostatic force of repulsion between them is,
$F_e=\frac{k\left(q\right)\left(q\right)}{d^2}=\frac{k{q}^2}{d^2}$
For $B$ to remain in equilibrium,
$N=mg\cos \theta$
and, $mg\sin \theta =F_e$
$\Rightarrow mg\sin {30}^{\circ }=\frac{k{q}^2}{d^2}$
$\Rightarrow d^2=\frac{k{q}^2}{mg\sin {30}^{\circ }}$
$\Rightarrow d^2=\frac{9\times {10}^9\times (2\times {10}^{-6}{)}^2}{\left(0.1\right)\left(10\right)\left(\frac{1}{2}\right)}$
$\Rightarrow d=\sqrt{\frac{72\times {10}^{-3}}{1}}$
$\therefore d=0.26836\text{m}\approx 27\text{cm}$
Thus, the distance of $B$ is approx $27\text{cm}$ from bottom.
Hence, option (d) is correct.

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: Draw the FBD of charge}{}^{\prime }{B}^{\prime }\text{and apply}\\ \text{equilibrium condition on it.}\end{array}$