A small oil drop of mass `10^(-6)` kg is hanging in at rest between two plates separated by 1 mm having in at rest between two plates separated by 1 mm having a potential difference of 500 V. The charge on the drop is `(g= 10 ms^(-2))`

Options

A. `2xx10^(-9) C`
B. `2xx10^(-11) C`
C. `2xx10^(-6) C`
D. `2xx10^(-9) C`

Answer

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

### Step 1: Identify the forces acting on the oil drop
The oil drop is acted upon by two forces:
1. The gravitational force (weight) acting downwards, given by \( F_g = mg \).
2. The electric force acting upwards, given by \( F_e = QE \).

### Step 2: Calculate the gravitational force
Given:
- Mass of the oil drop, \( m = 10^{-6} \) kg
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

The gravitational force is calculated as:
\[ F_g = mg = (10^{-6} \, \text{kg})(10 \, \text{m/s}^2) = 10^{-5} \, \text{N} \]

### Step 3: Calculate the electric field between the plates
The electric field \( E \) between two parallel plates is given by the formula:
\[ E = \frac{V}{d} \]
where:
- \( V = 500 \, \text{V} \) (potential difference)
- \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)

Substituting the values:
\[ E = \frac{500 \, \text{V}}{1 \times 10^{-3} \, \text{m}} = 500000 \, \text{V/m} = 5 \times 10^5 \, \text{V/m} \]

### Step 4: Set the electric force equal to the gravitational force
Since the drop is at rest, the net force acting on it is zero, which means:
\[ F_e = F_g \]
Thus, we have:
\[ QE = mg \]

### Step 5: Solve for the charge \( Q \)
Rearranging the equation gives:
\[ Q = \frac{mg}{E} \]

Substituting the values:
- \( mg = 10^{-5} \, \text{N} \)
- \( E = 5 \times 10^5 \, \text{V/m} \)

Calculating \( Q \):
\[ Q = \frac{10^{-5} \, \text{N}}{5 \times 10^5 \, \text{V/m}} = \frac{10^{-5}}{5 \times 10^5} = 2 \times 10^{-11} \, \text{C} \]

### Final Answer
The charge on the oil drop is \( Q = 2 \times 10^{-11} \, \text{C} \).

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