An ail drop having charge `2e` is kept stationary between two parallel horizontal plates `2.0 cm` apart when a potential difference of `12000` volts is applied between them. If the density of oil is `900 kg//m^(3)`, the radius of the drop will be

Options

A. `2.0 xx 10^(-6) m`
B. `1.7 xx 10^(-6) m`
C. `1.4xx 10^(-6) m`
D. `1.1 xx 10^(-6) m`

Explanation

To solve the problem, we need to find the radius of an oil drop with a charge of `2e` that is stationary between two parallel plates with a potential difference of `12000` volts and separated by `2.0 cm`. The density of the oil is given as `900 kg/m^3`.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Oil Drop**:
The oil drop is stationary, which means the net force acting on it is zero. The forces acting on the drop are:
- The electrostatic force due to the electric field between the plates.
- The gravitational force acting on the drop.

In equilibrium:
\[ F_{\text{electrostatic}} = F_{\text{gravity}} \]

2. **Calculate the Electric Field (E)**:
The electric field (E) between two parallel plates is given by the formula:
\[ E = \frac{V}{d} \]
where \( V \) is the potential difference and \( d \) is the distance between the plates.

Here, \( V = 12000 \, \text{volts} \) and \( d = 2.0 \, \text{cm} = 0.02 \, \text{m} \).

Substituting the values:
\[ E = \frac{12000}{0.02} = 600000 \, \text{V/m} \]

3. **Calculate the Electrostatic Force (F_e)**:
The electrostatic force on the oil drop can be calculated using:
\[ F_e = Q \cdot E \]
where \( Q = 2e \) and \( e \) (the charge of an electron) is approximately \( 1.6 \times 10^{-19} \, \text{C} \).

Therefore:
\[ Q = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \, \text{C} \]
Now substituting into the force equation:
\[ F_e = (3.2 \times 10^{-19}) \cdot (600000) = 1.92 \times 10^{-13} \, \text{N} \]

4. **Calculate the Gravitational Force (F_g)**:
The gravitational force acting on the 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 (approximately \( 10 \, \text{m/s}^2 \)).

The mass of the oil drop can be expressed in terms of its volume and density:
\[ m = \rho \cdot V = \rho \cdot \left(\frac{4}{3} \pi r^3\right) \]
where \( \rho = 900 \, \text{kg/m}^3 \).

Thus:
\[ F_g = \rho \cdot \left(\frac{4}{3} \pi r^3\right) \cdot g \]
Substituting the values:
\[ F_g = 900 \cdot \left(\frac{4}{3} \pi r^3\right) \cdot 10 \]
\[ F_g = 3000 \cdot \left(\frac{4}{3} \pi r^3\right) \]

5. **Set the Forces Equal**:
Since \( F_e = F_g \), we can set the two equations equal to each other:
\[ 1.92 \times 10^{-13} = 3000 \cdot \left(\frac{4}{3} \pi r^3\right) \]

6. **Solve for r**:
Rearranging the equation to solve for \( r^3 \):
\[ r^3 = \frac{1.92 \times 10^{-13}}{3000 \cdot \left(\frac{4}{3} \pi}\right)} \]
\[ r^3 = \frac{1.92 \times 10^{-13}}{4000 \pi} \]
Now calculating the value:
\[ r^3 = \frac{1.92 \times 10^{-13}}{12566.37} \approx 1.528 \times 10^{-17} \]
Taking the cube root:
\[ r \approx (1.528 \times 10^{-17})^{1/3} \approx 1.7 \times 10^{-6} \, \text{m} \]

### Final Answer:
The radius of the oil drop is approximately:
\[ r \approx 1.7 \, \mu m \]