1000 small water drops each of radius r and charge q coalesce together to form one spherical drop. The potential of the bigger drop is larger than that of the smaller one by a factor

Options

A. 1000
B. 100
C. 10
D. 1

Answer

To solve the problem, we need to determine how the potential of the larger drop compares to that of the smaller drops after they coalesce. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have 1000 small water drops, each with a radius \( r \) and charge \( q \). When these drops coalesce, they form a larger spherical drop. We need to find the factor by which the potential of the larger drop is greater than that of the smaller drops.

### Step 2: Calculate the Radius of the Larger Drop
The volume of the smaller drops combined must equal the volume of the larger drop. The volume \( V \) of a sphere is given by:

\[ V = \frac{4}{3} \pi r^3 \]

For 1000 small drops, the total volume is:

\[ V_{\text{total}} = 1000 \times \frac{4}{3} \pi r^3 \]

This equals the volume of the larger drop:

\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]

Setting these equal gives:

\[ 1000 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]

Cancelling \( \frac{4}{3} \pi \) from both sides:

\[ 1000 r^3 = R^3 \]

Taking the cube root:

\[ R = 10r \]

### Step 3: Calculate the Charge on the Larger Drop
Since the drops coalesce, the total charge on the larger drop \( Q \) is:

\[ Q = 1000q \]

### Step 4: Calculate the Capacitance of the Drops
The capacitance \( C \) of a spherical conductor is given by:

\[ C = 4 \pi \epsilon_0 r \]

For the smaller drops:

\[ C_{\text{small}} = 4 \pi \epsilon_0 r \]

For the larger drop:

\[ C_{\text{large}} = 4 \pi \epsilon_0 R = 4 \pi \epsilon_0 (10r) = 40 \pi \epsilon_0 r \]

### Step 5: Calculate the Potential of Each Drop
The potential \( V \) of a capacitor is given by:

\[ V = \frac{Q}{C} \]

For the smaller drops:

\[ V_{\text{small}} = \frac{q}{C_{\text{small}}} = \frac{q}{4 \pi \epsilon_0 r} \]

For the larger drop:

\[ V_{\text{large}} = \frac{Q}{C_{\text{large}}} = \frac{1000q}{40 \pi \epsilon_0 r} = \frac{25q}{\pi \epsilon_0 r} \]

### Step 6: Find the Ratio of the Potentials
Now, we find the ratio of the potential of the larger drop to that of the smaller drops:

\[ \frac{V_{\text{large}}}{V_{\text{small}}} = \frac{\frac{25q}{\pi \epsilon_0 r}}{\frac{q}{4 \pi \epsilon_0 r}} = \frac{25}{\frac{1}{4}} = 25 \times 4 = 100 \]

### Conclusion
The potential of the larger drop is larger than that of the smaller drops by a factor of **100**.

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