An oil drop having a charge q and apparent weight W falls with a terminal speed v in absence of electric field . When electric field is switched on it starts moving up with terminal speed v. The strength of electric field is

Options

A. `W/q`
B. `(2W)/q`
C. `(W)/(2q)`
D. `(4W)/q`

Answer

To solve the problem, we need to analyze the forces acting on the oil drop in two different scenarios: when there is no electric field and when there is an electric field.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Oil Drop (Without Electric Field)**:
- The oil drop has an apparent weight \( W \) acting downwards due to gravity.
- It experiences a viscous drag force \( F_{\text{viscous}} \) acting upwards when it falls with a terminal speed \( v \).
- At terminal velocity, these forces balance each other:
\[ F_{\text{viscous}} = W \]
- The viscous force can be expressed as:
\[ F_{\text{viscous}} = k v \]
- Thus, we have:
\[ k v = W \]

2. **Identify the Forces Acting on the Oil Drop (With Electric Field)**:
- When the electric field is switched on, the oil drop experiences an upward electric force \( F_{\text{electric}} \) in addition to the viscous drag.
- The electric force can be expressed as:
\[ F_{\text{electric}} = qE \]
- The forces acting on the drop now are:
- Weight \( W \) acting downwards
- Viscous force \( F_{\text{viscous}} = k v \) acting upwards
- Electric force \( F_{\text{electric}} = qE \) acting upwards
- At terminal velocity in this case, the forces also balance:
\[ F_{\text{electric}} + F_{\text{viscous}} = W \]
- Substituting the expressions for the forces:
\[ qE + k v = W \]

3. **Combine the Two Equations**:
- From the first scenario, we have:
\[ W = k v \]
- Substitute this into the second scenario's equation:
\[ qE + k v = k v \]
- Rearranging gives:
\[ qE = W - k v \]
- Since \( W = k v \), we can further simplify:
\[ qE = W - W = 0 \]
- This means that the electric force must equal the weight when the drop is moving upwards with terminal speed:
\[ qE = W + k v \]

4. **Solve for the Electric Field Strength \( E \)**:
- Rearranging the equation \( qE = W + k v \) gives:
\[ E = \frac{W + k v}{q} \]
- However, since we know \( k v = W \), we can substitute:
\[ E = \frac{W + W}{q} = \frac{2W}{q} \]

### Final Result:
The strength of the electric field \( E \) is:
\[ E = \frac{2W}{q} \]