A charged oil drop is suspended in uniform field of `3 xx 10^(4) V//m` so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge `= 9.9 xx 10^(-15)kg "&" g= 10 m//s^(2)`)

Options

A. `3.3 xx 10^(-18)C`
B. `3.2 xx 10^(-18)C`
C. `1.6 xx 10^(-18)C`
D. `4.8 xx 10^(-18)C`

Answer

To find the charge on the oil drop suspended in a uniform electric field, we can follow these steps:

### Step 1: Understand the forces acting on the oil drop
The oil drop is suspended in an electric field, which means the electric force acting on it is balanced by the gravitational force. Therefore, we can set up the equation:

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

### Step 2: Write the expressions for the forces
The electric force (\( F_{\text{electric}} \)) acting on the charged oil drop is given by:

\[ F_{\text{electric}} = q \cdot E \]

where \( q \) is the charge on the drop and \( E \) is the electric field strength.

The gravitational force (\( F_{\text{gravity}} \)) acting on the drop is given by:

\[ F_{\text{gravity}} = m \cdot g \]

where \( m \) is the mass of the drop and \( g \) is the acceleration due to gravity.

### Step 3: Set the forces equal to each other
Since the drop is in equilibrium (neither rising nor falling), we can set the two forces equal:

\[ q \cdot E = m \cdot g \]

### Step 4: Solve for the charge \( q \)
Rearranging the equation to solve for \( q \):

\[ q = \frac{m \cdot g}{E} \]

### Step 5: Substitute the known values
Now, we can substitute the given values into the equation:

- Mass of the drop, \( m = 9.9 \times 10^{-15} \, \text{kg} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Electric field strength, \( E = 3 \times 10^4 \, \text{V/m} \)

Substituting these values:

\[ q = \frac{(9.9 \times 10^{-15} \, \text{kg}) \cdot (10 \, \text{m/s}^2)}{3 \times 10^4 \, \text{V/m}} \]

### Step 6: Calculate the charge
Now, perform the calculation:

\[ q = \frac{9.9 \times 10^{-14} \, \text{kg} \cdot \text{m/s}^2}{3 \times 10^4 \, \text{V/m}} \]

\[ q = \frac{9.9 \times 10^{-14}}{3 \times 10^4} \]

\[ q = 3.3 \times 10^{-18} \, \text{C} \]

### Final Answer
The charge on the oil drop is:

\[ q = 3.3 \times 10^{-18} \, \text{C} \]

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