A vertical electric field of magnitude `4.9xx10^5` N/C just prevents a water droplet of a mass 0.1 g from falling. The value of charge on the droplet will be : (Given `g=9.8 m//s^2` )

Options

A. `1.6xx10^(-9)C`
B. `2.0xx10^(-9)C`
C. `3.2xx10^(-9)C`
D. `0.5xx10^(-9)C`

Answer

To find the charge on the water droplet that is just prevented from falling by a vertical electric field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Electric field strength, \( E = 4.9 \times 10^5 \, \text{N/C} \)
- Mass of the water droplet, \( m = 0.1 \, \text{g} = 0.1 \times 10^{-3} \, \text{kg} = 1.0 \times 10^{-4} \, \text{kg} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

2. **Calculate the Weight of the Droplet:**
The weight \( W \) of the droplet can be calculated using the formula:
\[ W = m \cdot g \]
Substituting the values:
\[ W = (1.0 \times 10^{-4} \, \text{kg}) \cdot (9.8 \, \text{m/s}^2) = 9.8 \times 10^{-4} \, \text{N} \]

3. **Understand the Forces Acting on the Droplet:**
The droplet is in equilibrium, meaning the upward electric force \( F_e \) due to the electric field must balance the downward weight of the droplet:
\[ F_e = W \]

4. **Relate Electric Force to Charge:**
The electric force \( F_e \) on a charge \( q \) in an electric field \( E \) is given by:
\[ F_e = q \cdot E \]
Setting this equal to the weight:
\[ q \cdot E = W \]

5. **Solve for Charge \( q \):**
Rearranging the equation to find \( q \):
\[ q = \frac{W}{E} \]
Substituting the values we have:
\[ q = \frac{9.8 \times 10^{-4} \, \text{N}}{4.9 \times 10^5 \, \text{N/C}} \]

6. **Calculate the Charge:**
Performing the calculation:
\[ q = \frac{9.8 \times 10^{-4}}{4.9 \times 10^5} = 2.0 \times 10^{-9} \, \text{C} \]

### Final Answer:
The value of the charge on the droplet is:
\[ \boxed{2.0 \times 10^{-9} \, \text{C}} \]