If 64 drps each charged to 220 V coalesce, what will be the potential of the bigger drop?

Answer

To solve the problem of finding the potential of a bigger drop formed by the coalescence of 64 smaller drops, each charged to 220 V, we can follow these steps:

### Step 1: Understand the relationship between the drops
When 64 smaller drops coalesce into one larger drop, the volume of the smaller drops will equal the volume of the larger drop. The volume of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the sphere.

### Step 2: Set up the volume equation
The total volume of the 64 smaller drops can be expressed as:
\[ 64 \times \frac{4}{3} \pi r^3 \]
The volume of the larger drop can be expressed as:
\[ \frac{4}{3} \pi R^3 \]
where \( R \) is the radius of the larger drop.

Setting these two volumes equal gives:
\[ 64 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]

### Step 3: Simplify the equation
We can cancel \( \frac{4}{3} \pi \) from both sides:
\[ 64 r^3 = R^3 \]

### Step 4: Solve for \( R \)
Taking the cube root of both sides, we find:
\[ R = 4r \]
This means the radius of the larger drop is 4 times the radius of the smaller drop.

### Step 5: Calculate the charge on each drop
The charge \( q \) on each smaller drop can be calculated using the formula:
\[ q = C \cdot V \]
where \( C \) is the capacitance. The capacitance of a sphere is given by:
\[ C = 4 \pi \epsilon_0 r \]
Thus, the charge on one smaller drop is:
\[ q = 4 \pi \epsilon_0 r \cdot 220 \]

### Step 6: Calculate the total charge on the larger drop
The total charge \( Q \) on the larger drop, which is the sum of the charges of all smaller drops, is:
\[ Q = 64q = 64 \cdot (4 \pi \epsilon_0 r \cdot 220) \]

### Step 7: Calculate the capacitance of the larger drop
The capacitance of the larger drop is:
\[ C' = 4 \pi \epsilon_0 R = 4 \pi \epsilon_0 (4r) = 16 \pi \epsilon_0 r \]

### Step 8: Calculate the potential of the larger drop
Using the formula for potential \( V' \) of the larger drop:
\[ V' = \frac{Q}{C'} \]
Substituting \( Q \) and \( C' \):
\[ V' = \frac{64 \cdot (4 \pi \epsilon_0 r \cdot 220)}{16 \pi \epsilon_0 r} \]

### Step 9: Simplify the expression
Canceling \( 4 \pi \epsilon_0 r \):
\[ V' = \frac{64 \cdot 220}{16} = 4 \cdot 220 = 880 \]

### Step 10: Final calculation
Since we have \( 64 \) drops, we multiply by \( 16 \):
\[ V' = 220 \cdot 16 = 3520 \]

Thus, the potential of the bigger drop is **3520 V**.

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