Charged Drops and Coalescence
For n identical isolated conducting drops of radius r and charge q coalescing without charge loss, R=n^(1/3)r and Q=nq before computing the new spherical potential.
Why this shows up in the exam
mercury-drop coalescence · potential scaling · electrostatic energy comparisons
Learn the idea
Coalescing identical drops conserves charge and volume, changing radius and potential. Many small conducting drops combine into one larger drop. Volumes add, charges add, and the larger radius grows only as the cube root of the number of drops.
🧠 Memory hook: Charge grows as n; radius grows as cube root n.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- R = n^(1/3) r — volume conservation for identical drops
- V_big/V_small = n^(2/3) — identical initial drops with conserved charge
How to approach it
- 1Use volume conservation for R
- 2Use charge conservation for Q
- 3Form V=kQ/R and simplify powers
Common slip-ups that cost marks
- •Taking R=nr
- •Holding potential fixed during isolated coalescence
- •Forgetting charge conservation
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
Two point charges 1 microC and 2 microC are 1 m apart in vacuum. Take k = 9 x 10^9 SI. Find the force magnitude.
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