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Step 1: Identify the Known Variables
• Number of identical mercury drops: 512
• Each drop’s potential: 2 V
• Let the charge on each small drop be q and its radius be r .
• The electrostatic potential of one small drop is given by V = \frac{kq}{r} , where k is the constant \frac{1}{4\pi \epsilon_0} .
Step 2: Express the Potential of Each Small Drop
From the given data, for each small drop:
2 = \frac{kq}{r}.
This tells us:
\frac{kq}{r} = 2.
Step 3: Relate the Radius of the Combined Drop to the Individual Drops
When the 512 drops coalesce to form a single large drop, their total volume must remain the same. Volume of one small drop is proportional to r^3 , and volume of the large drop is proportional to R^3 . Thus:
\frac{4}{3}\pi R^3 = 512 \times \frac{4}{3}\pi r^3.
Cancelling common factors:
R^3 = 512 \, r^3.
Therefore,
R = 8r.
Step 4: Sum of Charges on the Combined Drop
Each individual drop carries charge q . Hence, for 512 drops, total charge:
Q = 512 \, q.
Step 5: Compute the Potential of the Combined Drop
The potential of a spherical conductor is given by V_{\text{big}} = \frac{kQ}{R} . Substituting total charge Q = 512\,q and R = 8r :
V_{\text{big}} = \frac{k \, (512 \, q)}{8 \, r} = \frac{512}{8} \times \frac{kq}{r}.
Since \frac{kq}{r} = 2 from Step 2:
V_{\text{big}} = \frac{512}{8} \times 2 = 64 \times 2 = 128 \,\text{V}.
Answer: 128 V
