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Potential When Small Charged Drops Combine Into One

Physics · Electrostatic Potential And Capacitance · NEET

When n small identical charged drops combine into one big drop, the big drop's potential becomes V_big = n^(2/3) × v, where v is the potential of one small drop. This is because charge adds up (Q = nq) but the radius grows only as R = n^(1/3) × r. Memory hook: "charge grows fast, radius grows slow, so potential jumps by n to the power two-thirds."
27 small drops combine into one big dropn = 27 dropsradius r, potential v = 220 VmergeQ = 27qR = 3rV_big = 27^(2/3) v = 9 × 220R = n^(1/3) rQ = n qV_big = n^(2/3) v= 9 × 220 = 1980 V
27 identical small drops (radius r, potential 220 V) merge with charge conserved (Q = 27q) and volume conserved (R = 3r), so the big drop's potential is 27^(2/3) = 9 times larger = 1980 V.

Your doubts, answered

Why does the potential become n^(2/3) times and not n times?

Charge does add up n times (Q = nq), so you might expect potential to be n times bigger. But potential is V = kQ/R, and R also grows. Since volume is conserved, R = n^(1/3) × r. So V_big = k(nq) / (n^(1/3) r) = n / n^(1/3) × (kq/r) = n^(2/3) × v. The larger radius pulls the potential back down, giving n^(2/3), not n.

How do I find the radius of the big drop?

Use volume conservation. The liquid does not disappear, so the total volume of n small drops equals the volume of the one big drop: n × (4/3)π r^3 = (4/3)π R^3. Cancel (4/3)π: n r^3 = R^3, so R = n^(1/3) × r. For n = 27, R = 3r; for n = 8, R = 2r; for n = 1000, R = 10r.

Is charge conserved when drops combine?

Yes. Charge cannot be created or destroyed. If each small drop carries charge q, the big drop carries Q = nq. This is the key input, along with volume conservation, that gives the n^(2/3) result.

What about the electric field on the surface of the big drop?

Surface field is E = kQ/R^2 = k(nq)/(n^(1/3) r)^2 = n^(1/3) × (kq/r^2) = n^(1/3) × e. So the big drop's surface field is n^(1/3) times a small drop's field. Note: potential scales as n^(2/3) but surface field scales as n^(1/3) — a common NEET mix-up.

Does 27 drops really give 9 times the potential?

Yes. n^(2/3) with n = 27 is (27)^(2/3) = (3^3)^(2/3) = 3^2 = 9. So the big drop has 9 times the potential of one small drop. If each small drop was at 220 V, the big drop is at 1980 V (the NEET 2021 answer).

⚠️ The NEET trap
✗ Multiplying the small drop potential by n directly (27 × 220 = 5940 V), assuming potential scales with charge alone.
✓ Potential scales as n^(2/3) because the radius also grows as n^(1/3). For 27 drops: 27^(2/3) × 220 = 9 × 220 = 1980 V.
🧠 NTA loves this: charge is n times bigger, but the radius grows too, so potential is only n^(2/3) times. Never multiply by n.

Real NEET questions

NEET 2021

Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

A · 1520 V
B · 1980 V ✓
C · 660 V
D · 1320 V
Solution: Step 1 — Volume conservation: n small drops merge into one big drop, so n·(4/3)π r^3 = (4/3)π R^3, giving R = n^(1/3)·r. For n = 27, R = 27^(1/3)·r = 3r. Step 2 — Charge conservation: Q = nq = 27q. Step 3 — Potential: V_big = kQ/R = k(27q)/(3r) = 9·(kq/r) = 9·v. In general V_big = n^(2/3)·v = 27^(2/3)·v = 9·v. Step 4 — Put v = 220 V: V_big = 9 × 220 = 1980 V. Answer: (B) 1980 V.

Solved Electrostatic Potential And Capacitance NEET PYQs

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Frequently asked

What is the formula for potential of a big drop formed by n small drops?

V_big = n^(2/3) × v, where v is the potential of one small drop. This comes from charge Q = nq and radius R = n^(1/3) × r (volume conserved).

How does the radius change when n drops combine?

R = n^(1/3) × r. Volume is conserved, so n r^3 = R^3. For 8 drops R = 2r, for 27 drops R = 3r, for 64 drops R = 4r.

How does surface charge density change when drops combine?

Surface charge density σ = Q/(4πR^2) = nq/(4π (n^(1/3) r)^2) = n^(1/3) × σ_small. So it becomes n^(1/3) times the small drop's value.

Why is this topic important for NEET?

It is a direct-formula MCQ that appears often (e.g., NEET 2021). Once you know V scales as n^(2/3), you solve it in seconds. It also tests conservation of charge and volume together, which are core electrostatics ideas.

What is the potential energy relation for combined drops?

For a single charged sphere, U ∝ Q^2/R = (nq)^2/(n^(1/3) r) = n^(5/3) × (q^2/r). So the big drop's self-energy is n^(5/3) times a small drop's self-energy — a higher-order twist NEET occasionally uses.

Related concepts:
Potential and Surface Charge Density of a Charged ConductorPotential Inside and Outside a Charged Spherical ShellElectric Potential Due to a Point Charge (Derivation)Charge Sharing When Two Conductors Are ConnectedWhat Are Dielectrics and Polarisation?
Practise Physics for NEET:
Physics MCQs (free, chapter-wise)Electrostatic Potential And Capacitance PYQs