An oil drop, carrying six electronic charges and having a mass of `1.6xx10^(-12)g`, falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving down ward with ? Ignore buoyancy.

Options

A. `10^(5)NC^(-1)`
B. `10^(4)NC^(-1)`
C. `3.3xx10^(4)NC^(-1)`
D. `3.3xx10^(5)NC^(-1)`

Explanation

To solve the problem step by step, we will analyze the forces acting on the oil drop and use the relevant equations.

### Step 1: Understand the Forces Acting on the Oil Drop
When the oil drop is falling with terminal velocity, the forces acting on it are:
- The gravitational force (weight) acting downwards: \( F_g = mg \)
- The drag force acting upwards: \( F_d \)

At terminal velocity, these forces are equal:
\[ F_d = mg \]

### Step 2: Calculate the Weight of the Oil Drop
Given:
- Mass of the oil drop, \( m = 1.6 \times 10^{-12} \, \text{g} = 1.6 \times 10^{-15} \, \text{kg} \) (conversion from grams to kilograms)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

The weight of the oil drop is:
\[ F_g = mg = (1.6 \times 10^{-15} \, \text{kg})(9.8 \, \text{m/s}^2) \]
\[ F_g = 1.568 \times 10^{-14} \, \text{N} \]

### Step 3: Set Up the Equation for the Electric Field
When the electric field is applied to make the drop move upwards with the same speed, the forces acting on it are:
- The electric force acting upwards: \( F_e = Q \cdot E \)
- The drag force acting downwards: \( F_d \)
- The weight acting downwards: \( F_g \)

The equation of motion can be written as:
\[ F_e = F_d + F_g \]

Since the drop is moving upward with terminal velocity, we can express \( F_d \) as \( F_d = mg \) (because it balances the weight at terminal velocity):
\[ Q \cdot E = F_d + F_g \]
\[ Q \cdot E = mg + mg \]
\[ Q \cdot E = 2mg \]

### Step 4: Calculate the Charge on the Oil Drop
The oil drop carries 6 electronic charges. The charge of one electron is approximately \( e = 1.6 \times 10^{-19} \, \text{C} \).
Thus, the total charge \( Q \) is:
\[ Q = 6e = 6 \times (1.6 \times 10^{-19} \, \text{C}) = 9.6 \times 10^{-19} \, \text{C} \]

### Step 5: Substitute Values into the Electric Field Equation
Now substituting \( Q \) into the equation:
\[ E = \frac{2mg}{Q} \]

Substituting the values:
\[ E = \frac{2 \times (1.6 \times 10^{-15} \, \text{kg}) \times (9.8 \, \text{m/s}^2)}{9.6 \times 10^{-19} \, \text{C}} \]

### Step 6: Calculate the Electric Field
Calculating the numerator:
\[ 2mg = 2 \times 1.6 \times 10^{-15} \times 9.8 = 3.136 \times 10^{-14} \, \text{N} \]

Now substituting this into the equation for \( E \):
\[ E = \frac{3.136 \times 10^{-14}}{9.6 \times 10^{-19}} \]
\[ E \approx 3.27 \times 10^{4} \, \text{N/C} \]

### Step 7: Final Answer
Thus, the magnitude of the vertical electric field required to make the drop move upward with the same speed is approximately:
\[ E \approx 3.27 \times 10^{4} \, \text{N/C} \]