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 analyze the situation step by step.

**Step 1: Understanding the volume of the drops**

We know that the volume of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]

For 1000 small drops, each with radius \( r \), the total volume \( V_s \) is:
\[ V_s = 1000 \times \frac{4}{3} \pi r^3 = \frac{4000}{3} \pi r^3 \]

For the larger drop with radius \( R \), the volume \( V_L \) is:
\[ V_L = \frac{4}{3} \pi R^3 \]

Since the total volume remains the same when the small drops coalesce into one larger drop, we can set the two volumes equal:
\[ \frac{4000}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]

**Step 2: Simplifying the equation**

We can cancel out \( \frac{4}{3} \pi \) from both sides:
\[ 1000 r^3 = R^3 \]

Now, taking the cube root of both sides gives us:
\[ R = r \cdot 10 \]

**Step 3: Finding the capacitance of the drops**

The capacitance \( C \) of a spherical conductor is given by:
\[ C = 4 \pi \epsilon_0 r \]

For the smaller drop, the capacitance \( C_s \) is:
\[ C_s = 4 \pi \epsilon_0 r \]

For the larger drop, the capacitance \( C_L \) is:
\[ C_L = 4 \pi \epsilon_0 R = 4 \pi \epsilon_0 (10r) = 40 \pi \epsilon_0 r \]

**Step 4: Finding the ratio of capacitances**

Now, we can find the ratio of the capacitance of the larger drop to that of the smaller drop:
\[ \frac{C_L}{C_s} = \frac{40 \pi \epsilon_0 r}{4 \pi \epsilon_0 r} = 10 \]

**Step 5: Finding the potential of the drops**

The potential \( V \) of a charged conductor is given by:
\[ V = \frac{Q}{C} \]

For the smaller drop with charge \( q \):
\[ V_s = \frac{q}{C_s} = \frac{q}{4 \pi \epsilon_0 r} \]

For the larger drop with charge \( Q \) (which is the sum of the charges of the 1000 smaller drops, hence \( Q = 1000q \)):
\[ V_L = \frac{Q}{C_L} = \frac{1000q}{40 \pi \epsilon_0 r} = \frac{25q}{\pi \epsilon_0 r} \]

**Step 6: Finding the ratio of potentials**

Now we can find the ratio of the potentials:
\[ \frac{V_L}{V_s} = \frac{\frac{25q}{\pi \epsilon_0 r}}{\frac{q}{4 \pi \epsilon_0 r}} = \frac{25q \cdot 4 \pi \epsilon_0 r}{q \cdot \pi \epsilon_0 r} = 100 \]

Thus, the potential of the larger drop is larger than that of the smaller drop by a factor of **100**.

### Final Answer:
The potential of the bigger drop is larger than that of the smaller one by a factor of **100**.