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

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.