Surface Charge and Electrostatic Pressure
Just outside a conductor E_n = sigma/epsilon_0, and the mechanical pressure from the field is P = sigma^2/(2 epsilon_0) in vacuum.
Why this shows up in the exam
Charge density on irregular conductors · Forces joining charged hemispheres · Comparing concentric-shell surface densities
Learn the idea
Conductor charge crowds at sharper curvature, and its surface field produces an outward electrostatic pressure. A conductor is equipotential, so geometry determines how surface charge redistributes. Sharper regions generally need larger surface charge density to maintain the same potential.
🧠 Memory hook: Sharper surface, denser charge; pressure grows as sigma squared.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- E_out = sigma/epsilon₀ — normal field just outside a conductor
- P = sigma²/(2 epsilon₀) — electrostatic pressure on its surface
How to approach it
- 1Use equipotential geometry or symmetry
- 2Relate sigma to outside field
- 3Integrate pressure components over the required surface
Common slip-ups that cost marks
- •Assuming uniform sigma on every conductor
- •Using sigma/(2 epsilon₀) just outside a conductor
- •Multiplying pressure by the wrong projected area
🌟 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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