Differentiate the energy for the imposed electrical constraint
Track source work to reconcile fixed-charge and fixed-voltage forces.
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The question for today
What you’ll understand
- Derive force from a geometric energy variation at fixed charge.
- Account for electrical source work at fixed voltage.
- Use the appropriate transformed energy and audit force direction.
Electrostatic force can be found without adding every local charge force: change a geometric coordinate slightly and track energy. The method is compact, but its derivative must hold the correct electrical variable fixed. A disconnected capacitor stores fixed charge; a capacitor attached to an ideal voltage source exchanges charge and energy with that source. Ignoring the exchange makes the fixed-voltage calculation predict the wrong direction even though its formula for stored field energy is correct.
Choose a mechanical coordinate and a constraint
Let x describe a movable electrode position, with positive force defined along increasing x. In a quasistatic lossless variation, the electrical input V dQ is divided between stored electrostatic energy and mechanical work F dx. This accounting is local in the family of equilibrium configurations, not a claim that the apparatus moves with zero acceleration. Radiation, resistance, and material hysteresis require additional energy terms when they cannot be neglected.
At fixed charge there is no source input. The force is minus the partial derivative of stored energy with respect to position while charge is held fixed. For a linear capacitor the energy is Q²/(2C(x)), so increasing capacitance lowers the energy and attracts the mechanical coordinate toward larger capacitance. Define the coordinate before deciding its sign: increasing plate overlap increases capacitance, whereas increasing plate separation decreases it.
A voltage source pays an energy bill
When an ideal source fixes voltage, moving the electrode changes charge by voltage times the capacitance change. The source supplies V dQ, which is twice the change of field energy for a constant-voltage increment in capacitance. One half is stored in the field and the other half becomes mechanical work. The field energy can therefore rise while the electrostatic force drives the motion. Minimizing field energy alone is not the correct rule for this open electrical subsystem.
Subtract the source contribution by defining a transformed energy U−VQ. Its differential at fixed voltage is minus the mechanical work. For a linear capacitor it equals minus one half CV², yielding plus one half V² times the capacitance derivative for the force. This has the same instantaneous physical force as the fixed-charge expression after substituting Q=CV. The constraints define different neighboring states, not two competing force laws for the same state.
Audit signs with a parallel-plate limit
For plate separation d, capacitance is permittivity times area divided by d when fringing is negligible. Its derivative is negative, so the force component along increasing separation is negative: attraction. At fixed charge the magnitude is independent of separation within this ideal model. At fixed voltage it scales as inverse separation squared because the source changes the charge. The two protocols agree at the same initial state but follow different forces as separation changes.
The function lab isolates the derivative-sign step. Reset and set coefficient to 1 and position to 1; predict a slope of +2. Reset and set coefficient to −1 and position to 1; predict slope −2. A mechanical force from the displayed curve would have the opposite sign if the curve represented the appropriate energy. Its parabola is illustrative; it does not include capacitance, a voltage source, or a continuous charging protocol.
Work it through
An actuator gains capacitance
A linear actuator has C(x)=2 nF+(1 nF/mm)x over 0≤x≤1 mm and is held at 10 V. Find force along increasing x and the energy transfers for a small 0.10 mm displacement. Treat the capacitance slope as constant.
Convert the slope: 1 nF/mm equals 10⁻⁶ F/m. The constant-voltage force is ½×10²×10⁻⁶=5.0×10⁻⁵ N, positive because increasing x increases capacitance.
The motion increases capacitance by 0.10 nF and charge by VΔC=1.0 nC. The ideal source supplies VΔQ=10 nJ. This is the electrical work crossing into the capacitor-mechanical system.
The field energy increases by ½V²ΔC=5 nJ. The remaining 5 nJ is mechanical work, also FΔx=(5.0×10⁻⁵ N)(10⁻⁴ m). The two calculations agree, providing an explicit source-work audit.
The force is +5.0×10⁻⁵ N. The source supplies 10 nJ, divided into 5 nJ of added field energy and 5 nJ of mechanical work.
The living notebook
Explore the model
Interactive lab · function
Check the sign of an energy derivative
Reset and set a=1, x=1; predict slope +2 for y=ax². Reset and set a=−1, x=1; predict slope −2. A force derived from an appropriate potential would oppose that slope.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forThe lab is an adjustable parabola with its tangent and signed integral. It has no electrical source or capacitance, and changing controls does not implement a physical protocol.
Pause & reason
Check your understanding
Your turn
Try it yourself
At x=0 in the worked actuator, disconnect the source after charging to 10 V. Move to x=1 mm with charge fixed. Find final voltage and the external quasistatic work, neglecting losses.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Initial charge is C₀V₀. Use Q²/(2C) for both endpoint energies, not ½C times the initial voltage squared.
Compare with the solution
Initially Q=20 nC and U=100 nJ. The final capacitance is 3 nF, so final voltage is 20/3 V≈6.67 V and final energy is 66.67 nJ. External quasistatic work is ΔU=−33.33 nJ; the electrostatic system delivers +33.33 nJ mechanically. The force falls along this fixed-charge motion because this actuator’s capacitance increases.
Keep the idea
Take it with you
- Differentiate while holding the actual electrical control variable fixed.
- A fixed-voltage source exchanges energy when geometry changes.
- Fixed-charge and fixed-voltage formulas give the same force at the same state.
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