Question

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

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

Hint:

When combining drops, the potential scales with the radius, and charge conservation must be considered.

Answer 1

The Correct Option is C

Step 1: {Volume and charge conservation}
\[ 8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \Rightarrow R = 2r \] \[ 8q = Q \] Step 2: {Relating potentials}
\[ V' = \frac{kq}{r}, \quad V = \frac{kQ}{R} = \frac{k \times 8q}{2r} = 4 \frac{kq}{r} = 4V' \] \[ 20 = 4V' \Rightarrow V' = 5 \, {V} \] Thus, the potential of each single drop was 5 V.

Answer 2

Step 1: Formula for potential of a spherical conductor
The electric potential \( V \) of a spherical conductor (like a liquid drop) is given by:
\[ V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{R} \]
Where:
- \( Q \) is the charge on the drop
- \( R \) is the radius of the drop

Step 2: When identical drops combine
When \( n \) identical drops combine to form one large drop:
- The volume of the new drop becomes \( n \) times the volume of one small drop
- Since volume \( \propto R^3 \), the new radius \( R' \) becomes:
\[ R' = R \cdot n^{1/3} \]
- The total charge also becomes \( Q' = nQ \)

Step 3: Find relation between potential of big and small drops
Potential of small drop: \( V \propto \frac{Q}{R} \)
Potential of big drop: \( V' = \frac{nQ}{n^{1/3}R} = n^{2/3} \cdot \frac{Q}{R} = n^{2/3} V \)

Step 4: Apply given values
Given: \( n = 8 \), \( V' = 20\,V \)
So:
\[ V' = 8^{2/3} \cdot V = 4V \Rightarrow 20 = 4V \Rightarrow V = 5\,V \]

Final Answer: 5 V