The electric field strength in N `C^(-1)` that is required to just prevent a water drop carrying a change `1.6 xx 10^(-19)` C from falling under gravity is (g = 9.8 m `s^(-2)` , the mass of water drop = 0.0016 g )

Options

A. `9.8 xx10^(-16)`
B. `9.8 xx10^(16)`
C. `9.8 xx10^(-13)`
D. `9.8 xx10^(13)`

Answer

To solve the problem of finding the electric field strength required to prevent a water drop from falling under gravity, we can follow these steps:

### Step 1: Understand the Forces Acting on the Water Drop
The water drop has a charge \( Q = 1.6 \times 10^{-19} \, \text{C} \) and is subject to gravitational force \( F_g = mg \), where \( m \) is the mass of the drop and \( g \) is the acceleration due to gravity.

### Step 2: Convert Mass from Grams to Kilograms
The mass of the water drop is given as \( 0.0016 \, \text{g} \). We need to convert this to kilograms:
\[ m = 0.0016 \, \text{g} = 0.0016 \times 10^{-3} \, \text{kg} = 1.6 \times 10^{-6} \, \text{kg} \]

### Step 3: Calculate the Gravitational Force
Using the formula for gravitational force:
\[ F_g = mg \]
Substituting the values:
\[ F_g = (1.6 \times 10^{-6} \, \text{kg})(9.8 \, \text{m/s}^2) = 1.568 \times 10^{-5} \, \text{N} \]

### Step 4: Set Up the Equation for Electric Force
The electric force \( F_e \) acting on the charged water drop in an electric field \( E \) is given by:
\[ F_e = QE \]
To prevent the drop from falling, the electric force must balance the gravitational force:
\[ QE = mg \]

### Step 5: Solve for Electric Field Strength \( E \)
Rearranging the equation gives:
\[ E = \frac{mg}{Q} \]
Substituting the values we have:
\[ E = \frac{1.568 \times 10^{-5} \, \text{N}}{1.6 \times 10^{-19} \, \text{C}} \]

### Step 6: Calculate the Electric Field Strength
Now we perform the division:
\[ E = \frac{1.568 \times 10^{-5}}{1.6 \times 10^{-19}} = 9.8 \times 10^{13} \, \text{N/C} \]

### Final Answer
The electric field strength required to just prevent the water drop from falling is:
\[ E = 9.8 \times 10^{13} \, \text{N/C} \]

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