An oil drop carrying a charge q has a mass m kg. it is falling freely in air with terminal speed v. the electric field required to make, the drop move upwards with the same speed is

Options

A. `(mg)/(q)`
B. `(2mg)/(q)`
C. `(mgv)/(q^(2))`
D. `(2mgv)/(q)`

Explanation

To solve the problem, we need to analyze the forces acting on the oil drop when it is falling freely and when an electric field is applied to make it move upwards with the same terminal speed.

### Step-by-Step Solution:

1. **Understanding Terminal Velocity**:
When the oil drop is falling freely at terminal speed \( v \), the forces acting on it are:
- The gravitational force \( F_g = mg \) (downward)
- The drag force \( F_d \) (upward), which balances the gravitational force at terminal velocity.

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

2. **Applying Stokes' Law**:
According to Stokes' Law, the drag force \( F_d \) can be expressed as:
\[ F_d = 6 \pi \eta r v \]
where \( \eta \) is the viscosity of the fluid, \( r \) is the radius of the drop, and \( v \) is the terminal velocity.

Therefore, we can write:
\[ mg = 6 \pi \eta r v \quad \text{(1)} \]

3. **Considering the Electric Field**:
When an electric field \( E \) is applied to the oil drop, it experiences an electric force \( F_e \) given by:
\[ F_e = qE \]
where \( q \) is the charge of the oil drop.

4. **Forces When Electric Field is Applied**:
Now, when the electric field is applied, the forces acting on the drop are:
- The electric force \( F_e = qE \) (upward)
- The gravitational force \( F_g = mg \) (downward)
- The drag force \( F_d = 6 \pi \eta r v \) (upward)

The net force equation when the drop moves upwards with speed \( v \) is:
\[ qE - mg - F_d = 0 \]
Rearranging gives:
\[ qE = mg + F_d \quad \text{(2)} \]

5. **Substituting for Drag Force**:
From equation (1), we know that \( F_d = mg \). Substituting this into equation (2):
\[ qE = mg + mg \]
This simplifies to:
\[ qE = 2mg \]

6. **Finding the Electric Field**:
To find the electric field \( E \), we can rearrange the equation:
\[ E = \frac{2mg}{q} \]

### Final Answer:
The electric field required to make the drop move upwards with the same speed \( v \) is:
\[ E = \frac{2mg}{q} \]