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 solve the problem of finding the charge on the oil drop, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Forces Acting on the Oil Drop**:
The oil drop is in equilibrium, meaning the gravitational force acting downwards is balanced by the electric force acting upwards.

\[ F_{\text{gravity}} = mg \]
\[ F_{\text{electric}} = QE \]

2. **Set Up the Equation for Equilibrium**:
Since the drop is at rest, we can equate the two forces:
\[ mg = QE \]

3. **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 separation between the plates. Given \( V = 500 \, \text{V} \) and \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \):
\[ E = \frac{500 \, \text{V}}{1 \times 10^{-3} \, \text{m}} = 500000 \, \text{V/m} \]

4. **Substitute E into the Equilibrium Equation**:
Rearranging the equilibrium equation gives:
\[ Q = \frac{mg}{E} \]

5. **Substitute Known Values**:
We know:
- Mass \( m = 10^{-6} \, \text{kg} \)
- Gravitational acceleration \( g = 10 \, \text{m/s}^2 \)
- Electric field \( E = 500000 \, \text{V/m} \)

Now substituting these values into the equation:
\[ Q = \frac{(10^{-6} \, \text{kg})(10 \, \text{m/s}^2)}{500000 \, \text{V/m}} \]

6. **Calculate the Charge (Q)**:
\[ Q = \frac{10^{-5} \, \text{kg m/s}^2}{500000 \, \text{V/m}} = \frac{10^{-5}}{5 \times 10^{5}} = 2 \times 10^{-11} \, \text{C} \]

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

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