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Q.N identical small charged drops coalesce to form a large drop. Find the following ratio of large drop and small drop:

(i) Capacity
(ii) Potential
(iii) Charge
(iv) Electrostatic potential energy
Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2018Subjective· 2mImportance★★★★★
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Radius scales as N^(1/3); hence C ∝ N^(1/3), V ∝ N^(2/3), Q ∝ N, U ∝ N^(5/3).

When N identical small drops (each radius r, charge q) coalesce, volume is conserved:

N × (4/3)πr³ = (4/3)πR³ ⇒ R = N^(1/3) r.

Small drop: capacity c = 4πε₀r, charge q, potential v = q/(4πε₀r), energy u = ½ q v.

Big drop: charge Q = Nq, radius R = N^(1/3) r.

  1. Capacity: C = 4πε₀R = 4πε₀ N^(1/3) r ⇒ C/c = N^(1/3).
  2. Potential: V = Q/(4πε₀R) = Nq/(4πε₀ N^(1/3) r) = N^(2/3) × q/(4πε₀r) ⇒ V/v = N^(2/3). …

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