An oil drop of mass m fall through air with a terminal velocity in the presence of upward electric field of intensity `E`. the drop carries a charge `+q`. `R` is the viscous drag force and `f` is the buoyancy force . Them for the motion of the drop.

Options

A. there is a net force directed upward
B. there is a net force directed downward
C. `mg=Eq+F+R`
D. `mg+Eq=F+R`

Answer

To solve the problem of an oil drop falling through air with a terminal velocity in the presence of an upward electric field, we can analyze the forces acting on the drop.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Oil Drop**:
- The oil drop has a weight acting downward due to gravity, which is given by \( mg \), where \( m \) is the mass of the drop and \( g \) is the acceleration due to gravity.
- There is a viscous drag force \( R \) acting upward, which opposes the motion of the drop.
- The oil drop carries a charge \( +q \), and in the presence of an electric field \( E \), there is an electric force \( F_e \) acting upward, given by \( F_e = qE \).
- Additionally, there is a buoyant force \( F_b \) acting upward due to the displaced air.

2. **Set Up the Equation for Terminal Velocity**:
- At terminal velocity, the net force acting on the drop is zero because the forces are balanced. Therefore, we can write:
\[ mg = R + F_e + F_b \]
- Here, \( R \) is the viscous drag force, \( F_e \) is the electric force, and \( F_b \) is the buoyant force.

3. **Express the Viscous Drag Force**:
- The viscous drag force \( R \) can be expressed as:
\[ R = k v \]
- where \( k \) is a constant of proportionality and \( v \) is the terminal velocity.

4. **Substituting Forces into the Equation**:
- Substitute the expressions for the forces into the equation:
\[ mg = kv + qE + F_b \]

5. **Rearranging the Equation**:
- Rearranging the equation gives us:
\[ mg - qE - F_b = kv \]

6. **Conclusion**:
- The equation shows that at terminal velocity, the weight of the oil drop is balanced by the sum of the upward forces (viscous drag, electric force, and buoyancy). This balance of forces is crucial for understanding the motion of the drop in the electric field.