Two metal pieces having a potential difference of `800V` are `0.02 m` apart horizontally. A particle of mass `1.96xx10^(-15) kg` is suspended in equilibrium between the plates. If the `e` is the elementary charge, then charge on the particle is

Options

A. `e`
B. `3e`
C. `6e`
D. `8e`

Explanation

To find the charge on the particle suspended in equilibrium between two metal plates with a potential difference, we can follow these steps:

### Step 1: Understand the Forces Acting on the Particle
The particle is in equilibrium, meaning the electric force acting on it due to the electric field between the plates is balanced by the gravitational force acting on it. The gravitational force \( F_g \) can be expressed as:
\[ F_g = mg \]
where \( m \) is the mass of the particle and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

### Step 2: Calculate the Electric Field
The electric field \( E \) between two plates is given by the formula:
\[ E = \frac{V}{d} \]
where \( V \) is the potential difference and \( d \) is the distance between the plates. Given \( V = 800 \, \text{V} \) and \( d = 0.02 \, \text{m} \):
\[ E = \frac{800 \, \text{V}}{0.02 \, \text{m}} = 40000 \, \text{V/m} = 4 \times 10^4 \, \text{V/m} \]

### Step 3: Relate Electric Force to Charge
The electric force \( F_e \) acting on the charged particle is given by:
\[ F_e = qE \]
where \( q \) is the charge on the particle. Since the particle is in equilibrium, we can set the electric force equal to the gravitational force:
\[ qE = mg \]

### Step 4: Solve for Charge \( q \)
Rearranging the equation gives:
\[ q = \frac{mg}{E} \]
Substituting the known values:
- \( m = 1.96 \times 10^{-15} \, \text{kg} \)
- \( g = 9.8 \, \text{m/s}^2 \)
- \( E = 4 \times 10^4 \, \text{V/m} \)

Calculating \( mg \):
\[ mg = 1.96 \times 10^{-15} \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1.9208 \times 10^{-14} \, \text{N} \]

Now substituting into the equation for \( q \):
\[ q = \frac{1.9208 \times 10^{-14} \, \text{N}}{4 \times 10^4 \, \text{V/m}} = 4.802 \times 10^{-19} \, \text{C} \]

### Step 5: Express Charge in Terms of Elementary Charge \( e \)
The elementary charge \( e \) is approximately \( 1.6 \times 10^{-19} \, \text{C} \). To find the number of elementary charges \( n \) on the particle, we can express \( q \) as:
\[ q = n e \]
Thus,
\[ n = \frac{q}{e} = \frac{4.802 \times 10^{-19} \, \text{C}}{1.6 \times 10^{-19} \, \text{C}} \approx 3 \]

### Conclusion
The charge on the particle is:
\[ q = n e = 3 e \]