In Milikan's oil drop experiment, a charged drop falls with terminal velocity V. If an electric field E is applied in vertically upward direction then it starts moving in upward direction with terminal velocity 2V. If magnitude of electric field is decreased to `(E)/(2)`, then terminal velocity will become

Options

A. `(V)/(2)`
B. `V`
C. `(3V)/(2)`
D. 2V

Answer

To solve the problem, we will analyze the forces acting on the charged oil drop in Milikan's oil drop experiment under different electric field strengths and derive the terminal velocities accordingly.

### Step 1: Understand the forces acting on the oil drop
When the oil drop is falling under gravity, the forces acting on it are:
- The gravitational force \( F_g = mg \) acting downward.
- The viscous drag force \( F_d = 6 \pi R \eta V \) acting upward, where \( R \) is the radius of the drop, \( \eta \) is the viscosity of the fluid, and \( V \) is the terminal velocity.

At terminal velocity, these forces balance each other:
\[ mg = 6 \pi R \eta V \]

### Step 2: Analyze the situation with electric field \( E \)
When an electric field \( E \) is applied in the upward direction, the oil drop experiences an additional electric force \( F_e = qE \) acting upward, where \( q \) is the charge on the drop.

The new balance of forces when the drop moves upward with terminal velocity \( 2V \) is:
\[ mg = 6 \pi R \eta (2V) - qE \]

Rearranging gives:
\[ mg + qE = 12 \pi R \eta V \]

### Step 3: Relate the electric force to the gravitational force
From the first equation, we can express \( mg \) as:
\[ mg = 6 \pi R \eta V \]

Substituting this into the equation from Step 2:
\[ 6 \pi R \eta V + qE = 12 \pi R \eta V \]
\[ qE = 12 \pi R \eta V - 6 \pi R \eta V \]
\[ qE = 6 \pi R \eta V \]

### Step 4: Analyze the situation with electric field \( E/2 \)
Now, when the electric field is decreased to \( E/2 \), the new electric force becomes:
\[ F_e = q \left(\frac{E}{2}\right) = \frac{qE}{2} \]

The balance of forces at the new terminal velocity \( V' \) is:
\[ mg = 6 \pi R \eta V' - \frac{qE}{2} \]

Substituting \( qE = 6 \pi R \eta V \) from Step 3:
\[ mg = 6 \pi R \eta V' - \frac{1}{2}(6 \pi R \eta V) \]
\[ mg = 6 \pi R \eta V' - 3 \pi R \eta V \]

### Step 5: Substitute \( mg \) and solve for \( V' \)
Substituting \( mg = 6 \pi R \eta V \):
\[ 6 \pi R \eta V = 6 \pi R \eta V' - 3 \pi R \eta V \]
\[ 6 \pi R \eta V + 3 \pi R \eta V = 6 \pi R \eta V' \]
\[ 9 \pi R \eta V = 6 \pi R \eta V' \]

Dividing both sides by \( 6 \pi R \eta \):
\[ V' = \frac{9}{6} V = \frac{3}{2} V \]

### Conclusion
Thus, when the electric field is decreased to \( \frac{E}{2} \), the terminal velocity \( V' \) becomes:
\[ V' = \frac{3}{2} V \]