Question

A vertical electric field of magnitude \( 4.9 \times 10^5 \, {N/C} \) just prevents a water droplet of a mass \( 0.1 \, {g} \) from falling. The value of charge on the droplet will be: (Given \( g = 9.8 \, {m/s}^2 \))

• A. \( 1.6 \times 10^{-9} \, {C} \)
• B. \( 2.0 \times 10^{-9} \, {C} \)
• C. \( 3.2 \times 10^{-9} \, {C} \)
• D. \( 0.5 \times 10^{-9} \, {C} \)

Hint:

To find the charge on an object suspended in an electric field, equate the gravitational force to the electric force and solve for the charge.

Answer 1

The Correct Option is B

Step 1: The water droplet is at rest, so the upward electric force is equal to the downward gravitational force: \[ F_{{electric}} = F_{{gravity}}. \] 
Step 2: The electric force on the droplet is given by: \[ F_{{electric}} = qE, \] where \( q \) is the charge on the droplet and \( E = 4.9 \times 10^5 \, {N/C} \) is the electric field. 
Step 3: The gravitational force on the droplet is: \[ F_{{gravity}} = mg = 0.1 \times 10^{-3} \times 9.8 = 9.8 \times 10^{-4} \, {N}. \] 
Step 4: Setting the two forces equal: \[ qE = mg. \] Substituting the values: \[ q \times 4.9 \times 10^5 = 9.8 \times 10^{-4}. \] 
Step 5: Solving for \( q \): \[ q = \frac{9.8 \times 10^{-4}}{4.9 \times 10^5} = 2.0 \times 10^{-9} \, {C}. \]