Twenty seven drops of water of the same size are equally and similarly charged. They are then united to from a bigger drop. By what factor will the electrical potential changes.

Options

A. 9 times
B. 27 times
C. 6 times
D. 3 times

Explanation

To solve the problem of how the electrical potential changes when 27 equally charged drops of water combine to form a larger drop, we can follow these steps:

### Step 1: Understand the initial potential of the smaller drops
The potential \( V \) of a single small drop can be expressed as:
\[ V = \frac{k \cdot Q}{R} \]
where:
- \( k \) is Coulomb's constant,
- \( Q \) is the charge on the drop,
- \( R \) is the radius of the small drop.

### Step 2: Calculate the total charge of 27 small drops
If there are 27 drops, the total charge \( Q_{total} \) on these drops is:
\[ Q_{total} = 27Q \]

### Step 3: Calculate the potential due to all 27 drops
The potential \( V \) due to all 27 drops can be expressed as:
\[ V = \frac{k \cdot 27Q}{R} \]

### Step 4: Substitute the expression for charge \( Q \)
The charge \( Q \) can be related to the volume and density of the drops. The charge can be expressed as:
\[ Q = \rho \cdot V_{drop} = \rho \cdot \left(\frac{4}{3} \pi R^3\right) \]
Substituting this into the potential equation gives:
\[ V = \frac{k \cdot 27 \cdot \left(\rho \cdot \frac{4}{3} \pi R^3\right)}{R} \]
This simplifies to:
\[ V = 36k \pi R^2 \rho \]

### Step 5: Calculate the radius of the larger drop
When the 27 drops combine, the volume remains constant. Therefore, we have:
\[ 27 \cdot \left(\frac{4}{3} \pi R^3\right) = \frac{4}{3} \pi R_{big}^3 \]
From this, we can derive:
\[ R_{big}^3 = 27R^3 \implies R_{big} = 3R \]

### Step 6: Calculate the potential of the larger drop
Now, we can find the potential \( V' \) of the larger drop:
\[ V' = \frac{k \cdot Q_{big}}{R_{big}} \]
Where \( Q_{big} = 27Q \). Thus:
\[ V' = \frac{k \cdot 27Q}{3R} = \frac{27}{3} \cdot \frac{k \cdot Q}{R} = 9V \]

### Step 7: Determine the factor of change in potential
To find the factor by which the potential changes, we compare the potentials:
\[ \frac{V}{V'} = \frac{36k \pi R^2 \rho}{9V} = \frac{36}{9} = 4 \]

### Conclusion
Thus, the electrical potential changes by a factor of 4 when the 27 drops combine into one larger drop.