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)`

Answer

To solve the problem, we need to determine the electric field required to make an oil drop move upwards with the same terminal speed \( v \) that it has while falling freely under the influence of gravity and air resistance.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Oil Drop:**
- When the oil drop is falling freely, it experiences two forces:
- The gravitational force acting downwards: \( F_g = mg \)
- The viscous drag force acting upwards: \( F_d = 6 \pi \eta r v \) (according to Stokes' law)
- At terminal velocity, these forces balance out:
\[ mg = 6 \pi \eta r v \tag{1} \]

2. **Set Up the Forces When the Drop Moves Upwards:**
- When an electric field \( E \) is applied, it exerts an upward force on the charged drop:
- The electric force: \( F_e = qE \)
- The forces acting on the drop moving upwards are:
- The upward electric force: \( F_e = qE \)
- The downward gravitational force: \( F_g = mg \)
- The upward viscous drag force: \( F_d = 6 \pi \eta r v \)

3. **Balance the Forces for Upward Motion:**
- For the drop to move upwards with the same speed \( v \), the net force must be zero:
\[ qE = mg + 6 \pi \eta r v \tag{2} \]

4. **Substitute Equation (1) into Equation (2):**
- From Equation (1), we know that \( mg = 6 \pi \eta r v \). Therefore, we can substitute \( mg \) in Equation (2):
\[ qE = mg + mg \]
- This simplifies to:
\[ qE = 2mg \]

5. **Solve for the Electric Field \( E \):**
- Rearranging the equation gives:
\[ E = \frac{2mg}{q} \]

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