Question

Suppose a uniformly charged wall provides a uniform electric field of $2 \times 10^4 \, \text{N/C}$ normally. A charged particle of mass $2 \, \text{g}$ is suspended through a silk thread of length 20 cm and remains stayed at a distance of 10 cm from the wall. Then the charge on the particle will be \[\frac{1}{\sqrt{x}} \, \mu\text{C}, \, \text{where} \, x = \_.\]\text{[Use} $g = 10 \, \text{m/s}^2$\text{]}.

Answer 1

Correct Answer: 3

Given: - Electric field: \( E = 2 \times 10^4 \, \text{N/C} \) - Mass of the particle: \( m = 2 \, \text{g} = 2 \times 10^{-3} \, \text{kg} \) - Length of thread: \( L = 20 \, \text{cm} = 0.2 \, \text{m} \) - Distance of particle from the wall: \( d = 10 \, \text{cm} = 0.1 \, \text{m} \) - Acceleration due to gravity: \( g = 10 \, \text{m/s}^2 \)

Step 1: Analyzing the Forces Acting on the Particle

The charged particle experiences three forces: 1. Gravitational force (\( F_g = mg \)) 2. Tension (\( T \)) in the thread 3. Electric force (\( F_e = qE \))

For equilibrium, the components of forces must balance:

\[ F_e = T \sin\theta \quad \text{and} \quad F_g = T \cos\theta \]

where \( \theta \) is the angle between the thread and the vertical. From the geometry of the problem:

\[ \sin\theta = \frac{d}{L} = \frac{0.1}{0.2} = 0.5 \quad \text{and} \quad \cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - 0.5^2} = \sqrt{0.75} = \frac{\sqrt{3}}{2} \]

Step 2: Calculating the Tension

From the vertical equilibrium condition:

\[ T \cos\theta = mg \]

Substituting the values:

\[ T \times \frac{\sqrt{3}}{2} = 2 \times 10^{-3} \times 10 \] \[ T \times \frac{\sqrt{3}}{2} = 0.02 \] \[ T = \frac{0.02 \times 2}{\sqrt{3}} = \frac{0.04}{\sqrt{3}} \, \text{N} \]

Step 3: Calculating the Charge on the Particle

Using the horizontal equilibrium condition:

\[ F_e = T \sin\theta \] \[ qE = T \times 0.5 \]

Substituting the known values:

\[ q \times 2 \times 10^4 = \frac{0.04}{\sqrt{3}} \times 0.5 \] \[ q \times 2 \times 10^4 = \frac{0.02}{\sqrt{3}} \] \[ q = \frac{0.02}{\sqrt{3} \times 2 \times 10^4} \] \[ q = \frac{1}{\sqrt{3}} \times 10^{-6} \, \text{C} \]

Converting to microcoulombs:

\[ q = \frac{1}{\sqrt{3}} \, \mu\text{C} \]

Step 4: Comparing with the Given Expression

The charge is given as \( \frac{1}{\sqrt{x}} \, \mu\text{C} \). By comparison:

\[ x = 3 \]

Conclusion:

The value of \( x \) is 3.