A vertical electric field of magnitude `4.00xx 10^5 NC^(-1)` just prevents a water droplet of mass `1.000x 10^(-4) kg`from. falling, Find the charge on the droplet.

Options

A. `4.2 xx 10^-9 C`
B. `2.5 xx 10^-7 C`
C. `4.2 xx 10^-7 C`
D. `2.5 xx 10^-9 C`

Explanation

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

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Droplet**:
- The droplet experiences two main forces: the gravitational force (weight) acting downward and the electric force acting upward due to the electric field.

2. **Write the Expression for the Forces**:
- The gravitational force (weight) \( F_g \) acting on the droplet can be calculated using the formula:
\[ F_g = mg \]
where:
- \( m = 1.000 \times 10^{-4} \, \text{kg} \) (mass of the droplet)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

3. **Calculate the Gravitational Force**:
\[ F_g = (1.000 \times 10^{-4} \, \text{kg}) \times (9.8 \, \text{m/s}^2) = 9.8 \times 10^{-4} \, \text{N} \]

4. **Set Up the Equation for Electric Force**:
- The electric force \( F_e \) acting on the droplet due to the electric field \( E \) is given by:
\[ F_e = QE \]
where \( Q \) is the charge on the droplet and \( E = 4.00 \times 10^{5} \, \text{N/C} \) (magnitude of the electric field).

5. **Equate the Forces**:
- Since the droplet is in equilibrium and does not fall, the electric force must equal the gravitational force:
\[ QE = mg \]

6. **Solve for Charge \( Q \)**:
\[ Q = \frac{mg}{E} \]
Substituting the values we calculated:
\[ Q = \frac{(1.000 \times 10^{-4} \, \text{kg}) \times (9.8 \, \text{m/s}^2)}{4.00 \times 10^{5} \, \text{N/C}} \]

7. **Calculate the Charge**:
\[ Q = \frac{9.8 \times 10^{-4} \, \text{N}}{4.00 \times 10^{5} \, \text{N/C}} = 2.45 \times 10^{-9} \, \text{C} \]

### Final Answer:
The charge on the droplet is approximately \( 2.45 \times 10^{-9} \, \text{C} \).