Question

A particle A having a charge of $2.0 \times {10^{ - 6}}\;{\rm{C}}$ and a mass of $100\;{\rm{g}}$ 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 inclined plane so that B may remain in equilibrium?
A. $8\;{\rm{cm}}$from the bottom
B. $13\;{\rm{cm}}$from the bottom
C. $21\;{\rm{cm}}$from the bottom
D. $27\;{\rm{cm}}$from the bottom

Answer 1

The above problem can be resolved using the fundamentals of the equilibrium of forces. The equilibrium of force is being applied to this case successfully observed when the gravitational force and the electrostatic force becomes numerically equal, and the further variables can be solved accordingly.

Complete step by step answer:
Given:
The magnitude of charge of a particle is, $q = 2.0 \times {10^{ - 6}}\;{\rm{C}}$.
The mass of the particle is, $m = 100\;{\rm{g}} = 100\;{\rm{g}} \times \dfrac{{1\;{\rm{kg}}}}{{1000\;{\rm{g}}}} = 0.1\;{\rm{kg}}$.
The angle of inclination is, $\theta = 30\;^\circ$.
Let d be any distance of particle B from the bottom of the inclined plane. Then the force acting on the particle is,
${F_1} = mg\sin \theta$ ………………………………. (1)
Here, g is the gravitational acceleration and its value is $9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$.
The electrostatic force between A and B is,
${F_e} = \dfrac{{kq}}{{{d^2}}}$ ………………………………..(2)
Here, k is the coulomb’s constant and its magnitude is $9 \times {10^9}\;{\rm{N/C}}$.
Compare the equation 1 and 2 to satisfy the given equilibrium condition as,

{F_1} = {F_e}\\\ mg\sin \theta = \dfrac{{kq}}{{{r^2}}} \end{array}$$ Solve by substituting the values in above equation as, $$\begin{array}{l} mg\sin \theta = \dfrac{{k{q^2}}}{{{d^2}}}\\\ {d^2} = \dfrac{{9 \times {{10}^9}\;{\rm{N/C}} \times {{\left( {2 \times {{10}^{ - 6}}\;{\rm{C}}} \right)}^2}}}{{0.1\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \times \sin 30\;^\circ }}\\\ d = 0.27\;{\rm{m}} \end{array}$$ Convert the above result in cm as, $$d = 0.27\;{\rm{m}} = 0.27\;{\rm{m}} \times \dfrac{{100\;{\rm{cm}}}}{{1\;{\rm{m}}}} = 27\;{\rm{cm}}$$ Therefore, the particle B is at distance of $$27\;{\rm{cm}}$$ from the bottom and option (C) is correct. **Note:** To resolve the given problem, one must understand the meaning of equilibrium. Then this knowledge is being applied to solve the questions related to the equilibrium of forces. Along with this, the application of electrostatic force is to carry out, to analyse the interactions within the charges entities. Moreover, the knowledge of the distance between the entities is also necessary to remember.