Question

 A vertical electric field of magnitude 4.9 × 10⁵ N/C just prevents a water droplet of a mass of 0.1 g from falling. The value charge on the droplet will be(Given g = 9.8 m/s²)

• A. 1.6 × 10^–9 C
• B. 2.0 × 10^–9 C
• C. 3.2 × 10^–9 C
• D. 0.5 × 10^–9 C

Answer 1

The Correct Option is B

To determine the charge on the water droplet, we need to understand the equilibrium condition in which the electric field prevents the droplet from falling.

Given:

• Electric field, E = 4.9 \times 10^5 \, \text{N/C}
• Mass of the droplet, m = 0.1 \, \text{g} = 0.1 \times 10^{-3} \, \text{kg}
• Acceleration due to gravity, g = 9.8 \, \text{m/s}^2

For the droplet to remain stationary (i.e., not fall under gravity), the force due to the electric field must equal the gravitational force acting on it. Thus:

F_{\text{electric}} = F_{\text{gravity}}

Using the formula for electric force:

F_{\text{electric}} = q \cdot E

And for gravitational force:

F_{\text{gravity}} = m \cdot g

Equating the two forces:

q \cdot E = m \cdot g

We can solve for the charge q:

q = \frac{m \cdot g}{E}

Substitute the given values:

q = \frac{0.1 \times 10^{-3} \, \text{kg} \times 9.8 \, \text{m/s}^2}{4.9 \times 10^5 \, \text{N/C}}

Calculate the result:

q = \frac{0.00098 \, \text{kg}\cdot\text{m/s}^2}{490000 \, \text{N/C}}

q = 2.0 \times 10^{-9} \, \text{C}

Therefore, the charge on the water droplet is 2.0 × 10⁻⁹ C. This matches option 2 from the given choices, confirming it is the correct answer.