A charged oil drop falls with terminal velocity `v_(0)` in the absence of electric field. An electric field E keeps it stationary. The drop acquires charge 3q, it starts moving upwards with velocity `v_(0)`. The initial charge on the drop is

Options

A. `q/2`
B. q
C. `3/2q`
D. 2q

Answer

To solve the problem step by step, we will analyze the forces acting on the charged oil drop in different scenarios.

### Step 1: Analyze the first scenario (no electric field)
In the absence of an electric field, the charged oil drop falls with a terminal velocity \( v_0 \). At terminal velocity, the net force acting on the drop is zero. The forces acting on the drop are:
- Weight (\( mg \)) acting downward
- Drag force (\( F_D \)) acting upward

According to Stokes' law, the drag force can be expressed as:
\[ F_D = 6 \pi r \eta v_0 \]
Since the drop is falling at terminal velocity, we can equate the weight to the drag force:
\[ mg = 6 \pi r \eta v_0 \tag{1} \]

### Step 2: Analyze the second scenario (electric field keeps it stationary)
When an electric field \( E \) is applied, the drop is kept stationary. The forces acting on the drop now are:
- Weight (\( mg \)) acting downward
- Electric force (\( F_E \)) acting upward, given by \( F_E = EQ \) where \( Q \) is the charge on the drop.

Since the drop is stationary, these forces must balance:
\[ EQ = mg \tag{2} \]

### Step 3: Analyze the third scenario (drop acquires charge \( 3q \))
Now, the drop acquires a charge of \( 3q \) and starts moving upwards with velocity \( v_0 \). The forces acting on the drop are:
- Weight (\( mg \)) acting downward
- Drag force (\( F_D \)) acting downward, which is still \( 6 \pi r \eta v_0 \)
- Electric force (\( F_E \)) acting upward, now given by \( F_E = 3qE \)

Since the drop is moving upwards, we can write the equation of motion as:
\[ 3qE = mg + 6 \pi r \eta v_0 \tag{3} \]

### Step 4: Substitute \( mg \) from equation (2) into equation (3)
From equation (2), we have:
\[ mg = EQ \]
Substituting this into equation (3):
\[ 3qE = EQ + 6 \pi r \eta v_0 \]
Rearranging gives:
\[ 3qE - EQ = 6 \pi r \eta v_0 \]
Factoring out \( E \):
\[ (3q - Q)E = 6 \pi r \eta v_0 \]

### Step 5: Relate \( 6 \pi r \eta v_0 \) to \( mg \)
From equation (1), we know:
\[ mg = 6 \pi r \eta v_0 \]
Substituting this into the previous equation gives:
\[ (3q - Q)E = mg \]

### Step 6: Solve for the initial charge \( Q \)
Now, we can express \( mg \) in terms of \( EQ \) from equation (2):
\[ (3q - Q)E = EQ \]
Rearranging gives:
\[ 3qE = 2QE \]
Dividing both sides by \( E \) (assuming \( E \neq 0 \)):
\[ 3q = 2Q \]
Thus, solving for \( Q \):
\[ Q = \frac{3}{2}q \]

### Final Answer
The initial charge on the drop is:
\[ Q = \frac{3}{2}q \]