In Millike's oil drop experiment an oil drop carrying a charge `Q` is held stationary by a potential difference `2400 V` between the plates. To keep a drop of half the radius stationary the potential difference had to be made `600 V`. What is the charge on the second drop ?

Options

A. `(Q)/(4)`
B. `(Q)/(2)`
C. `Q`
D. `(3Q)/(2)`

Answer

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the relationship between charge, radius, and potential difference.
In Millikan's oil drop experiment, the charge \( Q \) on the oil drop is balanced by the electric force due to the electric field \( E \) created by the potential difference \( V \) between the plates and the gravitational force acting on the drop.

### Step 2: Write the formula for electric force and gravitational force.
The electric force \( F_e \) on the drop can be expressed as:
\[ F_e = Q \cdot E = Q \cdot \frac{V}{D} \]
where \( D \) is the distance between the plates.

The gravitational force \( F_g \) acting on the drop is given by:
\[ F_g = m \cdot g = \left( \frac{4}{3} \pi r^3 \rho \right) g \]
where \( r \) is the radius of the drop, \( \rho \) is the density of the oil, and \( g \) is the acceleration due to gravity.

### Step 3: Set the electric force equal to the gravitational force.
At equilibrium, these two forces are equal:
\[ Q \cdot \frac{V}{D} = \frac{4}{3} \pi r^3 \rho g \]

### Step 4: Rearrange the equation to express charge \( Q \).
From the above equation, we can express the charge \( Q \) as:
\[ Q = \frac{4}{3} \pi r^3 \rho g \cdot \frac{D}{V} \]

### Step 5: Establish the proportional relationship.
From the derived equation, we can see that:
\[ Q \propto \frac{r^3}{V} \]
This means that the charge \( Q \) is proportional to the cube of the radius \( r \) divided by the potential difference \( V \).

### Step 6: Set up the ratio for two drops.
Let \( Q_1 \) be the charge on the first drop with radius \( r_1 \) and potential difference \( V_1 \), and \( Q_2 \) be the charge on the second drop with radius \( r_2 \) and potential difference \( V_2 \). Then:
\[ \frac{Q_1}{Q_2} = \frac{r_1^3}{r_2^3} \cdot \frac{V_2}{V_1} \]

### Step 7: Substitute the known values.
Given:
- \( V_1 = 2400 \, V \)
- \( V_2 = 600 \, V \)
- \( r_1 = r \)
- \( r_2 = \frac{r}{2} \) (half the radius)

Substituting these values into the ratio:
\[ \frac{Q}{Q_2} = \frac{r^3}{\left(\frac{r}{2}\right)^3} \cdot \frac{600}{2400} \]

### Step 8: Simplify the equation.
Calculating the radius ratio:
\[ \frac{r^3}{\left(\frac{r}{2}\right)^3} = \frac{r^3}{\frac{r^3}{8}} = 8 \]

Now substituting this into the equation:
\[ \frac{Q}{Q_2} = 8 \cdot \frac{600}{2400} \]
\[ \frac{Q}{Q_2} = 8 \cdot \frac{1}{4} = 2 \]

### Step 9: Solve for \( Q_2 \).
Thus, we have:
\[ Q_2 = \frac{Q}{2} \]

### Final Answer:
The charge on the second drop is \( \frac{Q}{2} \).

---