A charged drop of mass `3.2 xx 10^(-12) g` floats between two horizontal parallel plates maintained at potential difference of 980 V and separation between the plates is 2 cm. The number of excess or deficient electrons on the drop is

Options

A. 2
B. 4
C. 8
D. 16

Explanation

To solve the problem of finding the number of excess or deficient electrons on a charged drop floating between two parallel plates, we can follow these steps:

### Step 1: Understand the given data
- Mass of the drop, \( m = 3.2 \times 10^{-12} \, \text{g} = 3.2 \times 10^{-15} \, \text{kg} \)
- Potential difference, \( \Delta V = 980 \, \text{V} \)
- Separation between the plates, \( D = 2 \, \text{cm} = 0.02 \, \text{m} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)
- Charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)

### Step 2: Calculate the electric field (\( E \))
The electric field between the plates is given by the formula:
\[ E = \frac{\Delta V}{D} \]
Substituting the values:
\[ E = \frac{980 \, \text{V}}{0.02 \, \text{m}} = 49000 \, \text{V/m} \]

### Step 3: Set up the force balance
The drop is in equilibrium, meaning the electric force acting on it is equal to its weight. The electric force (\( F_e \)) is given by:
\[ F_e = qE \]
Where \( q \) is the charge of the drop. The weight (\( W \)) of the drop is given by:
\[ W = mg \]
Since the drop is in equilibrium:
\[ qE = mg \]

### Step 4: Express the charge in terms of number of electrons
The charge \( q \) can be expressed as:
\[ q = n \cdot e \]
Where \( n \) is the number of excess or deficient electrons. Substituting this into the force balance equation gives:
\[ n \cdot e \cdot E = mg \]

### Step 5: Solve for \( n \)
Rearranging the equation to find \( n \):
\[ n = \frac{mg}{eE} \]
Substituting the known values:
\[ n = \frac{(3.2 \times 10^{-15} \, \text{kg}) \cdot (9.8 \, \text{m/s}^2)}{(1.6 \times 10^{-19} \, \text{C}) \cdot (49000 \, \text{V/m})} \]

### Step 6: Calculate \( n \)
Calculating the numerator:
\[ mg = 3.2 \times 10^{-15} \times 9.8 = 3.136 \times 10^{-14} \, \text{N} \]
Calculating the denominator:
\[ eE = 1.6 \times 10^{-19} \times 49000 = 7.84 \times 10^{-15} \, \text{N} \]
Now substituting these values into the equation for \( n \):
\[ n = \frac{3.136 \times 10^{-14}}{7.84 \times 10^{-15}} \approx 4 \]

### Conclusion
The number of excess or deficient electrons on the drop is approximately \( n = 4 \).

---