Two plane parallel conducting plates `1.5xx10^(-2)m` apart are held horizontal one above the other in air. The upper plate is maintained at a positive potential of`1.5kV` while the other plate is earthed. Calculate the number of electrons which must be attached to a small oil drop of mass `4.8xx10^(-15)kg` between the plates of move the drop with constant speed. Neglect the density and viscosity of air.

Explanation

To solve the problem, we need to calculate the number of electrons that must be attached to a small oil drop so that it moves with constant speed between two parallel conducting plates. Here are the steps to arrive at the solution:

### Step 1: Identify the Forces Acting on the Oil Drop
The oil drop experiences two main forces:
1. The gravitational force acting downward, \( F_g = mg \)
2. The electric force acting upward, \( F_e = qE \)

Where:
- \( m \) is the mass of the oil drop
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( q \) is the charge on the oil drop (which is the product of the number of electrons \( n \) and the charge of an electron \( e \))
- \( E \) is the electric field between the plates

### Step 2: 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 between the plates (in volts)
- \( d \) is the separation between the plates (in meters)

Given:
- \( V = 1.5 \, \text{kV} = 1500 \, \text{V} \)
- \( d = 1.5 \times 10^{-2} \, \text{m} \)

Substituting the values:
\[ E = \frac{1500 \, \text{V}}{1.5 \times 10^{-2} \, \text{m}} = 100000 \, \text{V/m} \]

### Step 3: Set Up the Equation for Constant Speed
For the oil drop to move with constant speed, the net force acting on it must be zero. Thus, we can equate the gravitational force to the electric force:
\[ mg = qE \]

Substituting \( q = ne \) (where \( e \) is the charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)):
\[ mg = neE \]

### Step 4: Solve for the Number of Electrons \( n \)
Rearranging the equation gives:
\[ n = \frac{mg}{eE} \]

### Step 5: Substitute the Known Values
Given:
- \( m = 4.8 \times 10^{-15} \, \text{kg} \)
- \( g = 9.81 \, \text{m/s}^2 \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
- \( E = 100000 \, \text{V/m} \)

Substituting these values into the equation:
\[ n = \frac{(4.8 \times 10^{-15} \, \text{kg})(9.81 \, \text{m/s}^2)}{(1.6 \times 10^{-19} \, \text{C})(100000 \, \text{V/m})} \]

Calculating the numerator:
\[ mg = 4.8 \times 10^{-15} \times 9.81 = 4.70688 \times 10^{-14} \, \text{N} \]

Calculating the denominator:
\[ eE = (1.6 \times 10^{-19})(100000) = 1.6 \times 10^{-14} \, \text{N} \]

Now substituting back:
\[ n = \frac{4.70688 \times 10^{-14}}{1.6 \times 10^{-14}} \approx 2.94 \approx 3 \]

### Final Answer:
The number of electrons that must be attached to the oil drop is approximately \( n = 3 \).

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