Question

A particle A having a charge of $2.0\times {10}^{-6}$C and a mass of 100 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?(Take g= $10m/s^2$)

• A. 8 cm from the bottom
• B. 13 cm from the bottom
• C. 21 cm from the bottom
• D. 27 cm from the bottom

Answer 1

The correct option is D 27 cm from the bottom

$\text{Step 1: Draw the free body diagram and Resolving forces[ Ref. Fig.]}$

$F_e$ is the Electrostatic force experienced between particle A and B.

The weight $mg$ of the particle can be resolved in two components as shown in the figure:

Along the incline plane: $mgsin\theta$

Perpendicular to the incline plane: $mgcos\theta$

where $\theta ={30}^o$

Let particle B be placed at a distance $x$ from particle A.

$\text{Step 2: Newton's Second Law}$

For particle B to remain in equilibrium the electrostatic force between the particles will be balanced by the component of the weight of particle B along incline direction i.e. acceleration $=0$

So, Applying Newton's Second Law along incline direction(Positive Upwards):

$\Sigma F=ma$

$F_e-mg\sin {30}^{\circ }=0$

$\Rightarrow F_e=mg\sin {30}^{\circ }$

$\Rightarrow \frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{x^2}=\frac{mg}{2}$ $....\left(1\right)$

$\text{Step 3: Substituting the values}$

Equation $\left(1\right)$

$\Rightarrow \frac{9\times {10}^9\times 2\times {10}^{-6}C\times 2\times {10}^{-6}C}{x^2}=\frac{0.1kg\times 10m/s^2}{2}$

$\Rightarrow \frac{36\times {10}^{-3}}{x^2}=\frac{1}{2}$

$\Rightarrow x^2=72\times {10}^{-3}m$

$\Rightarrow x=26.8\times {10}^{-2}m=27cm$

Hence, Option D is correct.