If n drops, each of capacitance C and charged to a potential V, coalesce to form a big drop, the ratio of the energy stored in the big drop to that in each small drop will be

Options

A. `n:1`
B. `n^(1/3):1`
C. `n^(5/3):1`
D. `n^2:1`

Answer

To solve the problem of finding the ratio of the energy stored in a big drop formed by n small drops to that in each small drop, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We have n small drops, each with capacitance \( C \) and charged to a potential \( V \). When these drops coalesce, they form a larger drop. We need to find the ratio of the energy stored in the big drop to that in each small drop.

2. **Energy Stored in a Capacitor**: The energy \( U \) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} \frac{Q^2}{C} \]
where \( Q \) is the charge and \( C \) is the capacitance.

3. **Charge on Each Small Drop**: The charge \( Q \) on each small drop can be expressed as:
\[ Q = CV \]

4. **Total Charge on n Small Drops**: The total charge \( Q_{\text{total}} \) when n small drops combine is:
\[ Q_{\text{total}} = nQ = nCV \]

5. **Energy Stored in Each Small Drop**: The energy stored in one small drop is:
\[ U_{\text{small}} = \frac{1}{2} \frac{(CV)^2}{C} = \frac{1}{2} CV^2 \]

6. **Energy Stored in All Small Drops**: The total energy stored in n small drops is:
\[ U_{\text{total small}} = n \cdot U_{\text{small}} = n \cdot \frac{1}{2} CV^2 = \frac{n}{2} CV^2 \]

7. **Capacitance of the Big Drop**: When the n small drops coalesce, the capacitance \( C' \) of the big drop can be calculated. The radius of the big drop \( R \) is related to the radius of the small drop \( r \) as:
\[ R = n^{1/3} r \]
The capacitance of a sphere is given by:
\[ C' = 4\pi \epsilon_0 R = 4\pi \epsilon_0 n^{1/3} r \]
Since \( C = 4\pi \epsilon_0 r \), we can express \( C' \) in terms of \( C \):
\[ C' = n^{1/3} C \]

8. **Energy Stored in the Big Drop**: The energy stored in the big drop is:
\[ U_{\text{big}} = \frac{1}{2} \frac{(Q_{\text{total}})^2}{C'} = \frac{1}{2} \frac{(nCV)^2}{n^{1/3}C} = \frac{1}{2} \frac{n^2C^2V^2}{n^{1/3}C} = \frac{1}{2} n^{5/3} CV^2 \]

9. **Finding the Ratio**: Now, we can find the ratio of the energy stored in the big drop to that in each small drop:
\[ \text{Ratio} = \frac{U_{\text{big}}}{U_{\text{small}}} = \frac{\frac{1}{2} n^{5/3} CV^2}{\frac{1}{2} CV^2} = n^{5/3} \]

### Final Result:
The ratio of the energy stored in the big drop to that in each small drop is:
\[ \text{Ratio} = n^{5/3} : 1 \]