A drop of `10^(-6)` kg water carries `10^(-6) C` charge. What electric field should be applied to balance its weight (assume `g = 10 ms^(-2)`)

Options

A. 10 V/m upward
B. 10 V/m downward
C. 0.1 V/m downward
D. 0.1 V/m upward

Answer

To solve the problem of determining the electric field required to balance the weight of a water droplet carrying a charge, we can follow these steps:

### Step 1: Identify the Given Values
- Mass of the water droplet, \( m = 10^{-6} \, \text{kg} \)
- Charge on the water droplet, \( q = 10^{-6} \, \text{C} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the Weight of the Water Droplet
The weight \( W \) of the water droplet can be calculated using the formula:
\[ W = m \cdot g \]
Substituting the values:
\[ W = 10^{-6} \, \text{kg} \cdot 10 \, \text{m/s}^2 = 10^{-5} \, \text{N} \]

### Step 3: Set Up the Equation for Electric Force
The electric force \( F_e \) experienced by the charged droplet in an electric field \( E \) is given by:
\[ F_e = q \cdot E \]
To balance the weight of the droplet, the electric force must equal the weight:
\[ q \cdot E = W \]

### Step 4: Substitute the Known Values into the Equation
Substituting the known values into the equation:
\[ 10^{-6} \, \text{C} \cdot E = 10^{-5} \, \text{N} \]

### Step 5: Solve for the Electric Field \( E \)
Rearranging the equation to solve for \( E \):
\[ E = \frac{10^{-5} \, \text{N}}{10^{-6} \, \text{C}} = 10 \, \text{V/m} \]

### Step 6: Conclusion
The electric field required to balance the weight of the water droplet is:
\[ E = 10 \, \text{V/m} \]
The direction of the electric field should be upward, as it needs to counteract the downward force of gravity on the positively charged droplet.