A charged water drop of weight `4 xx 10^(-18)` kg and charge `1.6 xx 10^(-19)` C is stable in an electric field . What is the intensity of electric field

Options

A. `150 V m^(-1)`
B. `200 V m^(-1)`
C. `250 V m^(-1)`
D. `50 V m^(-1)`

Answer

To find the intensity of the electric field in which a charged water drop is stable, we can follow these steps:

### Step 1: Understand the forces acting on the water droplet
The water droplet experiences two main forces:
1. The gravitational force (weight) acting downwards.
2. The electric force acting upwards due to the electric field.

### Step 2: Write down the equations for the forces
The weight of the droplet can be expressed as:
\[ W = m \cdot g \]
where:
- \( m = 4 \times 10^{-18} \) kg (mass of the droplet)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)

The electric force acting on the droplet can be expressed as:
\[ F_e = q \cdot E \]
where:
- \( q = 1.6 \times 10^{-19} \) C (charge of the droplet)
- \( E \) is the electric field intensity we need to find.

### Step 3: Set the forces equal for equilibrium
For the droplet to be stable (in equilibrium), the upward electric force must equal the downward gravitational force:
\[ F_e = W \]
Thus, we have:
\[ q \cdot E = m \cdot g \]

### Step 4: Substitute the known values
Substituting the values into the equation:
\[ (1.6 \times 10^{-19}) \cdot E = (4 \times 10^{-18}) \cdot (9.81) \]

### Step 5: Calculate the weight of the droplet
Calculating the weight:
\[ W = 4 \times 10^{-18} \cdot 9.81 = 3.924 \times 10^{-17} \, \text{N} \]

### Step 6: Solve for the electric field \( E \)
Now, we can rearrange the equation to solve for \( E \):
\[ E = \frac{W}{q} = \frac{3.924 \times 10^{-17}}{1.6 \times 10^{-19}} \]

### Step 7: Perform the division
Calculating the electric field:
\[ E = \frac{3.924 \times 10^{-17}}{1.6 \times 10^{-19}} = 24525 \, \text{V/m} \]

### Step 8: Round to appropriate significant figures
Rounding this to two significant figures gives:
\[ E \approx 25000 \, \text{V/m} \]

### Final Answer
The intensity of the electric field is approximately:
\[ E \approx 25000 \, \text{V/m} \]