In Millikan's experiment, an oil drop of radius `10^(-4)` cm remains suspended between the plates which are 1 cm apart. If the drop has charge of 5e over it, calculate the potential difference between the plates. The density of oil may be taken as `1.5 gcm^(-3)`.

Answer

To solve the problem step by step, we will follow the logic presented in the video transcript while ensuring clarity and accuracy.

### Step 1: Understand the Given Data
- Radius of the oil drop, \( r = 10^{-4} \) cm = \( 10^{-6} \) m
- Charge on the oil drop, \( Q = 5e = 5 \times 1.6 \times 10^{-19} \) C
- Distance between the plates, \( d = 1 \) cm = \( 10^{-2} \) m
- Density of oil, \( \rho = 1.5 \) g/cm³ = \( 1.5 \times 10^{3} \) kg/m³

### 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 \( r = 10^{-6} \) m:
\[ V = \frac{4}{3} \pi (10^{-6})^3 = \frac{4}{3} \pi \times 10^{-18} \text{ m}^3 \]

### Step 3: Calculate the Mass of the Oil Drop
Using the density to find the mass \( m \):
\[ m = \rho \times V = \rho \times \frac{4}{3} \pi r^3 \]
Substituting the values:
\[ m = 1.5 \times 10^{3} \times \frac{4}{3} \pi (10^{-6})^3 \]
Calculating this gives:
\[ m = 1.5 \times 10^{3} \times \frac{4}{3} \pi \times 10^{-18} = 2 \pi \times 10^{-15} \text{ kg} \]

### Step 4: Calculate the Gravitational Force on the Oil Drop
The gravitational force \( F_g \) acting on the drop is given by:
\[ F_g = mg \]
Using \( g \approx 9.8 \text{ m/s}^2 \):
\[ F_g = (2 \pi \times 10^{-15}) \times 9.8 \]

### Step 5: Set Up the Equation for Electric Force
The electric force \( F_e \) acting on the drop due to the electric field \( E \) is given by:
\[ F_e = QE \]
Where \( E = \frac{V}{d} \). Therefore,
\[ F_e = Q \frac{V}{d} \]

### Step 6: Equate the Forces
Since the drop is suspended, the gravitational force equals the electric force:
\[ mg = QE \implies mg = Q \frac{V}{d} \]
Rearranging for \( V \):
\[ V = \frac{mgd}{Q} \]

### Step 7: Substitute Values to Find \( V \)
Substituting \( m \), \( g \), \( d \), and \( Q \):
\[ Q = 5 \times 1.6 \times 10^{-19} = 8 \times 10^{-19} \text{ C} \]
Now substituting:
\[ V = \frac{(2 \pi \times 10^{-15}) \times 9.8 \times (10^{-2})}{8 \times 10^{-19}} \]

### Step 8: Calculate the Final Value of \( V \)
Calculating the above expression:
\[ V = \frac{(2 \pi \times 9.8 \times 10^{-17})}{8 \times 10^{-19}} = \frac{(19.6 \pi \times 10^{-17})}{8 \times 10^{-19}} = \frac{19.6 \pi}{8} \times 10^{2} \]
Using \( \pi \approx 3.14 \):
\[ V \approx \frac{19.6 \times 3.14}{8} \times 10^{2} \approx 7.69 \times 10^{2} \text{ V} \approx 769.6 \text{ V} \]

### Final Answer
The potential difference between the plates is approximately \( 770 \text{ V} \).

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