A charged oil drop falls terminal velocity `V_0` in the absence of electric field . An electric field E keeps it stationary . The drop acquires additional charge q and starts moving upwards with velocity `V_(0)`. The intial charge on the drop was .

Options

A. 4q
B. 2q
C. q
D. q/2

Explanation

To solve the problem step by step, we need to analyze the forces acting on the charged oil drop in two scenarios: when it is falling with terminal velocity and when it is kept stationary by an electric field after acquiring additional charge.

### Step 1: Analyze the falling oil drop
When the oil drop is falling with terminal velocity \( V_0 \) in the absence of an electric field, the forces acting on it are:
- The gravitational force \( mg \) acting downward.
- The viscous drag force \( F_d \) acting upward, which can be expressed using Stokes' law as:
\[ F_d = 6 \pi \eta r V_0 \]

At terminal velocity, these forces are balanced:
\[ mg = 6 \pi \eta r V_0 \tag{1} \]

### Step 2: Introduce the electric field
When an electric field \( E \) is applied, the drop acquires an additional charge \( q \) and starts moving upwards with the same velocity \( V_0 \). In this case, the forces acting on the drop are:
- The gravitational force \( mg \) acting downward.
- The viscous drag force \( F_d \) acting downward, which remains the same as before:
\[ F_d = 6 \pi \eta r V_0 \]
- The electric force \( F_e \) acting upward, given by:
\[ F_e = qE \]

At this point, the forces are also balanced, so we can write:
\[ F_e = mg + F_d \]
Substituting the expressions for the forces, we get:
\[ qE = mg + 6 \pi \eta r V_0 \tag{2} \]

### Step 3: Substitute for \( mg \) from Step 1
From equation (1), we have:
\[ mg = 6 \pi \eta r V_0 \]
Substituting this into equation (2):
\[ qE = 6 \pi \eta r V_0 + 6 \pi \eta r V_0 \]
\[ qE = 2(6 \pi \eta r V_0) \]
\[ qE = 12 \pi \eta r V_0 \]

### Step 4: Solve for the initial charge \( Q \)
Now, we need to find the initial charge \( Q \) on the drop. We know that the drop was initially falling with terminal velocity \( V_0 \) and had a charge \( Q \). We can relate the charge \( Q \) to the forces acting on it in the electric field:
\[ QE = mg + 6 \pi \eta r V_0 \]
Substituting \( mg \) from equation (1):
\[ QE = 6 \pi \eta r V_0 + 6 \pi \eta r V_0 \]
\[ QE = 12 \pi \eta r V_0 \]

### Step 5: Relate \( q \) and \( Q \)
From the previous equations, we can see that:
\[ q = 2Q \]
Thus, the initial charge \( Q \) can be expressed as:
\[ Q = \frac{q}{2} \]

### Final Answer
The initial charge on the drop was \( \frac{q}{2} \).

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