There are n electrons of charge on a drop of oil of density `rho`. It is in equilibrium in an electric field E. Then radius of drop is

Options

A. `[(2n e E)/(4pi rho g)]^(1//3) `
B. `[(n eE)/(rho g)]`
C. `[(3n e E)/(4pi rho g)]^(1//3)`
D. `[(2n e E)/(pi rho g)]`

Answer

To find the radius of the oil drop in equilibrium in an electric field, we can follow these steps:

### Step 1: Understand the forces acting on the oil drop
In equilibrium, the electric force acting on the oil drop is balanced by the gravitational force. The electric force \( F_e \) is given by:
\[ F_e = q \cdot E \]
where \( q \) is the total charge on the drop and \( E \) is the electric field strength.

### Step 2: Calculate the total charge on the oil drop
The total charge \( q \) on the oil drop can be expressed in terms of the number of electrons \( n \) and the charge of an electron \( e \):
\[ q = n \cdot e \]

### Step 3: Write the expression for the gravitational force
The gravitational force \( F_g \) acting on the oil drop is given by:
\[ F_g = m \cdot g \]
where \( m \) is the mass of the oil drop and \( g \) is the acceleration due to gravity.

### Step 4: Calculate the mass of the oil drop
The mass \( m \) can be calculated using the density \( \rho \) and the volume \( V \) of the drop. The volume of a spherical drop is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Thus, the mass can be expressed as:
\[ m = \rho \cdot V = \rho \cdot \frac{4}{3} \pi r^3 \]

### Step 5: Set the electric force equal to the gravitational force
Since the drop is in equilibrium, we can set the electric force equal to the gravitational force:
\[ q \cdot E = m \cdot g \]
Substituting the expressions for \( q \) and \( m \):
\[ (n \cdot e) \cdot E = \left(\rho \cdot \frac{4}{3} \pi r^3\right) \cdot g \]

### Step 6: Rearranging the equation
Rearranging the equation gives:
\[ n \cdot e \cdot E = \rho \cdot \frac{4}{3} \pi r^3 \cdot g \]
Now, we can solve for \( r^3 \):
\[ r^3 = \frac{3 n e E}{4 \pi \rho g} \]

### Step 7: Taking the cube root to find the radius
Taking the cube root of both sides, we find the radius \( r \):
\[ r = \sqrt[3]{\frac{3 n e E}{4 \pi \rho g}} \]

### Final Answer
Thus, the radius of the oil drop is:
\[ r = \sqrt[3]{\frac{3 n e E}{4 \pi \rho g}} \]