A charged oil drop of mass `2.5xx10^(-7)`kg is in space between the two plates, each of area `2xx10^(-2) m^(2)` of a parallel plate capacitor. When the upper plate has a charge of `5 xx10^(-7) C` and the lower plate has an equal negative charge, then the oil remains stationary. the charge of the oil drop is (take, `g=10 m//s^(2)`)

Options

A. `9xx10^(-1) C`
B. `9xx10^(-6) C`
C. `8.85xx10^(-13) C`
D. `1.8xx10^(-14) C`

Answer

To find the charge of the oil drop, we can follow these steps:

### Step 1: Understand the forces acting on the oil drop
The oil drop is subjected to two forces:
1. The gravitational force acting downwards, given by \( F_g = mg \).
2. The electric force acting upwards due to the electric field between the plates, given by \( F_e = Q \cdot E \).

### Step 2: Set up the equation for equilibrium
Since the oil drop remains stationary, the upward electric force must balance the downward gravitational force:
\[ F_e = F_g \]
This can be expressed as:
\[ Q \cdot E = mg \]

### Step 3: Calculate the electric field (E) between the plates
The electric field \( E \) between two parallel plates with charge \( Q \) and area \( A \) is given by:
\[ E = \frac{\sigma}{\epsilon_0} = \frac{Q/A}{\epsilon_0} = \frac{Q}{A \cdot \epsilon_0} \]
where \( \sigma \) is the surface charge density and \( \epsilon_0 \) is the permittivity of free space.

### Step 4: Substitute the expression for E into the equilibrium equation
Substituting \( E \) into the equilibrium equation gives:
\[ Q \cdot \left(\frac{Q}{A \cdot \epsilon_0}\right) = mg \]
This simplifies to:
\[ \frac{Q^2}{A \cdot \epsilon_0} = mg \]

### Step 5: Solve for Q
Rearranging the equation to solve for \( Q \):
\[ Q^2 = mg \cdot A \cdot \epsilon_0 \]
\[ Q = \sqrt{mg \cdot A \cdot \epsilon_0} \]

### Step 6: Substitute the known values
Given:
- Mass \( m = 2.5 \times 10^{-7} \) kg
- Gravitational acceleration \( g = 10 \, \text{m/s}^2 \)
- Area \( A = 2 \times 10^{-2} \, \text{m}^2 \)
- Permittivity of free space \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)

Substituting these values into the equation:
\[ Q = \sqrt{(2.5 \times 10^{-7} \, \text{kg}) \cdot (10 \, \text{m/s}^2) \cdot (2 \times 10^{-2} \, \text{m}^2) \cdot (8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2)} \]

### Step 7: Calculate the value
Calculating the expression inside the square root:
\[ Q = \sqrt{(2.5 \times 10^{-7}) \cdot (10) \cdot (2 \times 10^{-2}) \cdot (8.85 \times 10^{-12})} \]
\[ = \sqrt{(2.5 \times 10^{-7} \cdot 10 \cdot 2 \cdot 8.85 \times 10^{-12})} \]
\[ = \sqrt{(5.425 \times 10^{-18})} \]
\[ = 7.37 \times 10^{-9} \, \text{C} \]

### Final Answer
The charge of the oil drop is approximately \( 7.37 \times 10^{-9} \, \text{C} \).