Electric Potential: Practice Problem & Solution
1000 identical drops of mercury are charged to a potential of \(1 \text{V}\) each. They coalesce to form a single drop, the potential of this new drop will be
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Electric Potential. Understanding the underlying formula is key to arriving at the correct answer below:
The potential of a coalesced drop is related to the individual potential by \(V' = n^{2/3} V\). For \(n = 1000\), \(V' = (1000)^{2/3} \times 1 = 100 \text{V}\).
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