A Charged particle of mass 1.0 g is suspended through a
. silk thread of length 40 cm in a horizantal electric field
. of `4.0 xx 10 ^4 NC ^(-1)` . If the particle stays aty a distance of
. 24 cm from the wall in equilibrium, find the charge on
. the particle.

Options

A. `0.9 xx 10^(-7) C`
B. `1.8 xx 10^(-7) C`
C. `2.7 xx 10^(-7) C`
D. `0.9 xx 10^(-9) C`

Explanation

To find the charge on the particle suspended in a horizontal electric field, we will follow these steps:

### Step 1: Understand the Forces Acting on the Charged Particle
The charged particle is in equilibrium, meaning the net force acting on it is zero. The forces acting on the particle are:
- The gravitational force (weight) acting downward, \( W = mg \).
- The electric force acting horizontally, \( F_e = qE \).

### Step 2: Calculate the Weight of the Particle
Given:
- Mass of the particle, \( m = 1.0 \, \text{g} = 1.0 \times 10^{-3} \, \text{kg} \)
- Acceleration due to gravity, \( g \approx 9.81 \, \text{m/s}^2 \)

The weight \( W \) can be calculated as:
\[ W = mg = (1.0 \times 10^{-3} \, \text{kg})(9.81 \, \text{m/s}^2) = 9.81 \times 10^{-3} \, \text{N} \]

### Step 3: Identify the Geometry of the Setup
The length of the silk thread is \( L = 40 \, \text{cm} = 0.4 \, \text{m} \). The particle is at a distance of \( 24 \, \text{cm} = 0.24 \, \text{m} \) from the wall.

Using the Pythagorean theorem, we can find the vertical distance \( h \) from the point of suspension to the horizontal line of the electric field:
\[ h = \sqrt{L^2 - d^2} = \sqrt{(0.4)^2 - (0.24)^2} = \sqrt{0.16 - 0.0576} = \sqrt{0.1024} = 0.32 \, \text{m} \]

### Step 4: Calculate the Angles
Using the geometry, we can find \( \sin \theta \) and \( \cos \theta \):
\[ \sin \theta = \frac{d}{L} = \frac{0.24}{0.4} = 0.6 \]
\[ \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - (0.6)^2} = \sqrt{0.64} = 0.8 \]

### Step 5: Set Up the Equilibrium Condition
In equilibrium, the vertical component of the tension \( T \) balances the weight, and the horizontal component of the tension balances the electric force:
\[ T \sin \theta = W \quad \text{(1)} \]
\[ T \cos \theta = qE \quad \text{(2)} \]

### Step 6: Divide the Two Equations
Dividing equation (1) by equation (2):
\[ \frac{T \sin \theta}{T \cos \theta} = \frac{W}{qE} \]
This simplifies to:
\[ \tan \theta = \frac{W}{qE} \]
Rearranging gives:
\[ q = \frac{W}{E \tan \theta} \]

### Step 7: Calculate \( \tan \theta \)
Using the values of \( \sin \theta \) and \( \cos \theta \):
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{0.6}{0.8} = 0.75 \]

### Step 8: Substitute Values to Find Charge \( q \)
Substituting \( W \), \( E \), and \( \tan \theta \):
\[ q = \frac{9.81 \times 10^{-3} \, \text{N}}{(4.0 \times 10^4 \, \text{N/C})(0.75)} \]
Calculating:
\[ q = \frac{9.81 \times 10^{-3}}{3.0 \times 10^4} = 3.27 \times 10^{-7} \, \text{C} \]

### Final Answer
The charge on the particle is approximately:
\[ q \approx 3.27 \times 10^{-7} \, \text{C} \]