Force on a Movable Dielectric
For a quasistatic coordinate x, the electrical force follows the derivative of the appropriate total energy: at fixed Q, F=-d[Q^2/(2C)]/dx; with an ideal voltage source, F=(1/2)V^2 dC/dx after source work is included.
Why this shows up in the exam
dielectric slab pull-in · liquid rise in a capacitor · variable-overlap actuators
Learn the idea
A dielectric is pulled toward the position that increases capacitance. Inserting dielectric lets the system store more charge at fixed voltage or lower field energy at fixed charge; either full energy accounting predicts an inward force.
🧠 Memory hook: The system pulls toward larger C.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- F = (1/2)V² dC/dx — fixed V with ideal source
- F = (Q²/(2C²)) dC/dx — fixed Q
How to approach it
- 1Express C as a function of displacement
- 2State fixed Q or V
- 3Differentiate and balance other forces
Common slip-ups that cost marks
- •Using the same energy derivative without checking the constraint
- •Ignoring gravity in liquid-rise equilibrium
- •Using total slab area instead of changing overlap
🌟 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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