An oil drop of radiuss `2mm` with density `3(g/(cm^3))` is held stationary under a constant `vec(E) = 3.35 xx 10^5 V/m` in the milikan's drop experiment. What is number of excess electron that oil drop will possess `(g=9.81)m/s^2`

Answer

To solve the problem of determining the number of excess electrons an oil drop possesses in the Millikan oil drop experiment, we will follow these steps:

### Step 1: Convert Given Values to SI Units
- **Radius of the oil drop**: Given as `2 mm`, we convert this to meters:
\[ r = 2 \text{ mm} = 2 \times 10^{-3} \text{ m} \]
- **Density of the oil drop**: Given as `3 g/cm³`, we convert this to kg/m³:
\[ \text{Density} = 3 \text{ g/cm}^3 = 3 \times 10^{3} \text{ kg/m}^3 \]
- **Electric field**: Given as `3.35 × 10^5 V/m`, this value remains the same.

### Step 2: Calculate the Volume of the Oil Drop
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting the value of \( r \):
\[ V = \frac{4}{3} \pi (2 \times 10^{-3})^3 \]
Calculating \( V \):
\[ V = \frac{4}{3} \pi (8 \times 10^{-9}) = \frac{32\pi}{3} \times 10^{-9} \text{ m}^3 \]

### Step 3: Calculate the Mass of the Oil Drop
Using the formula for mass \( m = \text{Density} \times V \):
\[ m = 3 \times 10^{3} \text{ kg/m}^3 \times \frac{32\pi}{3} \times 10^{-9} \text{ m}^3 \]
Calculating \( m \):
\[ m = 32\pi \times 10^{-6} \text{ kg} \]

### Step 4: Calculate the Weight of the Oil Drop
The weight \( W \) of the oil drop is given by:
\[ W = mg \]
Substituting the values:
\[ W = 32\pi \times 10^{-6} \times 9.81 \text{ N} \]

### Step 5: Relate Weight to Electric Force
The electric force \( F \) acting on the oil drop is given by:
\[ F = qE \]
Where \( q \) is the charge on the oil drop and \( E \) is the electric field strength. Since the oil drop is stationary, the electric force equals the weight:
\[ qE = mg \]

### Step 6: Solve for Charge \( q \)
Rearranging the equation gives:
\[ q = \frac{mg}{E} \]
Substituting the values:
\[ q = \frac{32\pi \times 10^{-6} \times 9.81}{3.35 \times 10^{5}} \]

### Step 7: Calculate the Number of Excess Electrons
The charge \( q \) can also be expressed in terms of the number of excess electrons \( n \):
\[ q = n \cdot e \]
Where \( e = 1.6 \times 10^{-19} \text{ C} \) (the charge of an electron). Thus:
\[ n = \frac{q}{e} \]
Substituting for \( q \):
\[ n = \frac{\frac{32\pi \times 10^{-6} \times 9.81}{3.35 \times 10^{5}}}{1.6 \times 10^{-19}} \]

### Step 8: Final Calculation
Calculating \( n \) will give us the number of excess electrons.