Question

As shown in the figure, a configuration of two equal point charges ($q_0 = +2\mu C$) is placed on an inclined plane. Mass of each point charge is $20g$. Assume that there is no friction between charge and plane. For the system of two point charges to be in equilibrium (at rest) the height $h = x \times 10^{-3}m$ The value of $x$ is ____ mm. (Take $\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9} Nm^{2}C^{-2}, g = 10ms^{-1}$)

Answer 1

300

Answer 2

To solve this problem, we need to analyze the forces acting on the upper point charge ($q_0$) to ensure the system is in equilibrium. Since there is no friction, only gravitational force and electrostatic force are relevant along the inclined plane.

1. Identify the forces:

   • Gravitational force (mg): Acts vertically downwards.
   • Electrostatic force ($F_e$): Both charges are positive, so they repel each other. The electrostatic force on the upper charge acts upwards along the inclined plane, away from the lower charge.

2. Resolve gravitational force: The component of gravitational force acting down the inclined plane is $mg \sin \theta$.

3. Equilibrium condition: For the system to be in equilibrium (at rest), the electrostatic repulsive force must balance the component of gravitational force acting down the incline. $F_e = mg \sin \theta$

4. Express electrostatic force: The electrostatic force between two point charges is given by Coulomb's Law: $F_e = k \frac{q_0^2}{r^2}$ where $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ Nm}^2\text{C}^{-2}$, $q_0$ is the charge, and $r$ is the distance between the two charges.

5. Relate distance 'r' to height 'h': From the figure, 'h' is the vertical height, and 'r' is the distance along the inclined plane between the two charges. The angle of inclination is $\theta = 30^\circ$. Using trigonometry, $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{r}$. Therefore, $r = \frac{h}{\sin \theta}$.

6. Substitute and solve for 'h': Substitute the expressions for $F_e$ and $r$ into the equilibrium equation: $k \frac{q_0^2}{r^2} = mg \sin \theta$ $k \frac{q_0^2}{(h/\sin \theta)^2} = mg \sin \theta$ $k \frac{q_0^2 \sin^2 \theta}{h^2} = mg \sin \theta$ Assuming $\sin \theta \neq 0$ (which is true for $\theta = 30^\circ$), we can cancel one $\sin \theta$ term: $k \frac{q_0^2 \sin \theta}{h^2} = mg$ Now, solve for $h^2$: $h^2 = \frac{k q_0^2 \sin \theta}{mg}$ $h = \sqrt{\frac{k q_0^2 \sin \theta}{mg}}$

7. Plug in the given values:

   • $q_0 = +2\mu C = 2 \times 10^{-6} C$
   • $m = 20g = 20 \times 10^{-3} kg = 0.02 kg$
   • $k = 9 \times 10^9 \text{ Nm}^2\text{C}^{-2}$
   • $g = 10 \text{ ms}^{-2}$
   • $\theta = 30^\circ \implies \sin \theta = \sin 30^\circ = 0.5$

   $h = \sqrt{\frac{(9 \times 10^9) \times (2 \times 10^{-6})^2 \times 0.5}{(0.02) \times 10}}$ $h = \sqrt{\frac{9 \times 10^9 \times (4 \times 10^{-12}) \times 0.5}{0.2}}$ $h = \sqrt{\frac{36 \times 10^{-3} \times 0.5}{0.2}}$ $h = \sqrt{\frac{18 \times 10^{-3}}{0.2}}$ $h = \sqrt{\frac{18 \times 10^{-3}}{2 \times 10^{-1}}}$ $h = \sqrt{9 \times 10^{-2}}$ $h = 3 \times 10^{-1} m$ $h = 0.3 m$

8. Determine the value of x: The problem states that $h = x \times 10^{-3} m$. We found $h = 0.3 m$. To match the format, convert $0.3 m$ to $10^{-3} m$: $0.3 m = 0.3 \times 1000 \times 10^{-3} m = 300 \times 10^{-3} m$. Comparing this with $h = x \times 10^{-3} m$, we get $x = 300$.

The question asks for the value of $x$ in mm. This phrasing is ambiguous because $x$ is defined as a dimensionless number in $h = x \times 10^{-3}m$. However, it is common in such questions that the final numerical value is expected. If the question implicitly means "the value of $h$ in mm is $x$", then $h = 0.3 m = 300 mm$, so $x=300$. In either interpretation, the numerical value of $x$ is 300.