The potential of a large liquid drop when eight liquid drops are combined is 20 V. Then, the potential of each single drop was

Options

A. 10 V
B. 7.5 V
C. 5 V
D. 2.5 V

Answer

To find the potential of each single drop when eight liquid drops are combined to form a larger drop with a potential of 20 V, we can follow these steps:

### Step 1: Understand the potential of a single drop
The potential \( V \) of a single drop (small sphere) with charge \( Q \) and radius \( r \) is given by the formula:
\[ V = \frac{kQ}{r} \]
where \( k \) is the electrostatic constant.

### Step 2: Combine the drops
When eight such drops are combined, the total charge of the larger drop becomes:
\[ Q_{total} = 8Q \]
The radius of the larger drop \( R \) can be determined from the volume conservation principle. The volume of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Thus, the volume of eight small drops is:
\[ V_{small} = 8 \times \frac{4}{3} \pi r^3 = \frac{32}{3} \pi r^3 \]
The volume of the larger drop is:
\[ V_{large} = \frac{4}{3} \pi R^3 \]
Setting these equal gives:
\[ \frac{32}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
Cancelling \( \frac{4}{3} \pi \) from both sides:
\[ 32r^3 = R^3 \]
Taking the cube root:
\[ R = 2r \]

### Step 3: Calculate the potential of the larger drop
The potential \( V_B \) of the larger drop is given as 20 V. The potential of the larger drop can also be expressed as:
\[ V_B = \frac{kQ_{total}}{R} = \frac{k(8Q)}{R} \]
Substituting \( R = 2r \):
\[ V_B = \frac{k(8Q)}{2r} = 4 \frac{kQ}{r} \]

### Step 4: Relate the potentials
From the earlier equation for the potential of a single drop:
\[ V = \frac{kQ}{r} \]
We can express the potential of the larger drop in terms of the potential of a single drop:
\[ V_B = 4V \]

### Step 5: Solve for the potential of a single drop
Given \( V_B = 20 \) V, we can find \( V \):
\[ 20 = 4V \implies V = \frac{20}{4} = 5 \text{ V} \]

### Final Answer
Thus, the potential of each single drop is:
\[ \boxed{5 \text{ V}} \]