Sixty four conducting drops each of radius 0.02 m and each carrying a charge of 5 μC are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be :

Solution

Physics sir · 3 steps

Volume is shared when drops merge, so the radius grows only as the cube root while the charge grows 64 times.

1. Find the new radius. The total volume stays the same: 4/3π R³ = 64 × 4/3π r³. So R³ = 64r³, which gives R = 4r.
2. Find the new charge. Charge is conserved, so the big drop carries Q = 64q. On a conductor this charge sits on the outer surface, so σ = Q/(4π R²).

   σ_(big)/σₛₘₐₗₗ = (64q / 4π (4r)²)/(q / 4π r²) = 64/16 = 4

3. So. The bigger drop has 4 times the surface charge density, a ratio of 4 : 1. The radius 0.02 m and the charge 5 μC cancel out and are not needed for the ratio.

The idea to keep. When n identical drops merge, R = n^(1/3)r and Q = nq. Then σ scales as n^(1/3), potential as n^(2/3) and capacitance as n^(1/3).

Answer from the NTA final answer key. Solution by PrepWiser.