A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength `(81pi)/(7)xx10^5Vm^-1`. When the field is switched off, the drop is observed to fall with terminal velocity `2xx10^-3ms^-1`. Given `g=9.8ms^-2`, viscoisty of the air `=1.8xx10^-5Nsm^-2` and the denisty of oil `=900kg m^-3`, the magnitude of q is

Options

A. `1.6 xx 10^(-19)` C
B. `3.2 xx 10^(-19)` C
C. `4.8 xx 10^(-19)` C
D. `8.0 xx 10^(-19)` C

Explanation

To find the magnitude of the charge \( q \) on the oil drop, we can follow these steps:

### Step 1: Write down the given data
- Electric field strength, \( E = \frac{81\pi}{7} \times 10^5 \, \text{V/m} \)
- Terminal velocity, \( V = 2 \times 10^{-3} \, \text{m/s} \)
- Viscosity of air, \( \eta = 1.8 \times 10^{-5} \, \text{N s/m}^2 \)
- Density of oil, \( \rho = 900 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Use the terminal velocity formula
When the electric field is switched off, the oil drop falls with a terminal velocity \( V \). The formula for terminal velocity \( V \) of a sphere in a viscous fluid is given by:

\[ V = \frac{2R^2(\rho - \sigma)g}{9\eta} \]

Where:
- \( R \) is the radius of the drop,
- \( \sigma \) is the density of air (approximately \( 1.2 \, \text{kg/m}^3 \) for air).

### Step 3: Rearranging the formula to find \( R \)
Rearranging the formula to solve for \( R \):

\[ R^2 = \frac{9V\eta}{2(\rho - \sigma)g} \]

### Step 4: Substitute the known values to find \( R \)
Substituting the known values into the equation:

\[ R^2 = \frac{9 \times (2 \times 10^{-3}) \times (1.8 \times 10^{-5})}{2 \times (900 - 1.2) \times 9.8} \]

Calculating the denominator:

\[ 900 - 1.2 = 898.8 \]

Now substituting:

\[ R^2 = \frac{9 \times 2 \times 10^{-3} \times 1.8 \times 10^{-5}}{2 \times 898.8 \times 9.8} \]

Calculating \( R^2 \) gives us the radius \( R \).

### Step 5: Calculate the charge \( q \)
The charge \( q \) can be calculated using the formula:

\[ q = \frac{4}{3} \pi R^3 \rho g \]

And using the electric field strength \( E \):

\[ q = \frac{1}{E} \left( \frac{4}{3} \pi R^3 \rho g \right) \]

### Step 6: Substitute \( R \) and \( E \) to find \( q \)
Substituting the calculated value of \( R \) and the value of \( E \):

\[ q = \frac{1}{\frac{81\pi}{7} \times 10^5} \left( \frac{4}{3} \pi R^3 \cdot 900 \cdot 9.8 \right) \]

### Step 7: Final Calculation
After substituting all values and simplifying, you will find:

\[ q \approx 8 \times 10^{-19} \, \text{C} \]

### Conclusion
Thus, the magnitude of the charge \( q \) on the oil drop is approximately \( 8 \times 10^{-19} \, \text{C} \).

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