Ask "does voltage lead or lag the current in a capacitor?" and the recurring wrong answer is "voltage leads by 90 degrees." It's the inductor's answer, swapped onto the capacitor. The reason it's tempting is that ELI and ICE blur together: under pressure you remember that one of them has voltage out in front, you just can't recall which, so you reach for the half of the mnemonic you happen to remember and land on the mirror image. For a capacitor the correct answer is the opposite: current leads voltage by 90 degrees.
The fix isn't a better mnemonic. It's to stop memorizing the sign at all and derive it from the defining equation in one line. The derivative always pushes the quantity it acts on forward by 90 degrees, and that single fact rebuilds the whole thing under exam pressure when the mnemonic has gone fuzzy. ICE is the capacitor (I before C, with E in the middle); ELI is the inductor (E before I, with L in the middle) — but you'll see below you never have to trust which is which.
This is one of the most common places people invert a sign. On the All About Circuits forum there's a long ELI-the-ICE-man thread where people keep re-deriving the result to convince themselves which way it goes, and Quora has repeated versions of the same question. The pattern is always the same: the mnemonic is a memory crutch with no derivation attached, so once it fades, there's nothing to fall back on and the guess is a coin flip.
Why people flip it
Look at the impedances. For an inductor, Z_L = jωL. For a capacitor, Z_C = 1/(jωC) = -j/(ωC). The j and the -j look like bookkeeping — arbitrary symbols you carry around. Nothing in "j" tells your gut that voltage is ahead of current, so when the exam asks for the imaginary part's sign the j feels like a thing to recall rather than reconstruct.
"Lead" and "lag" make it worse because they're relative. Lead with respect to what? If you don't fix a reference, "voltage leads" and "current lags" describe the same situation, and it's easy to talk yourself into the mirror image.
The one line that fixes it
The constitutive laws are the anchor. An inductor obeys v = L·di/dt. A capacitor obeys i = C·dv/dt. Each says one quantity is the derivative of the other (times a constant). And the derivative of cos(ωt) is -ω·sin(ωt), where -sin(ωt) = cos(ωt + 90°). So differentiating a cosine shifts it forward by 90 degrees — a quarter cycle. That's the whole secret.
Apply it to the inductor. Let the current be i = I·cos(ωt). Then
v = L·di/dt = L · (-ω·I·sin(ωt)) = L·ω·I·cos(ωt + 90°).
The voltage is the differentiated quantity, so it's the one shoved forward: voltage at +90 degrees, current at 0. Voltage leads current. That is ELI.
Now the capacitor. Let v = V·cos(ωt). Then
i = C·dv/dt = C · (-ω·V·sin(ωt)) = C·ω·V·cos(ωt + 90°).
This time the current is the differentiated quantity, so current is at +90 and voltage at 0. Current leads voltage. That is ICE.
You never memorized "voltage leads in an inductor." You wrote down v = L·di/dt, differentiated a cosine, and read the sign off the +90 the derivative produced. Same procedure both times; only the quantity on the derivative changes.
A fully worked example you can check
Take ω = 1000 rad/s.
Inductor. Let L = 10 mH = 0.01 H, driven by i(t) = 2·cos(1000t) A. Then
v = L·di/dt = 0.01 · d/dt[2·cos(1000t)] = 0.01 · (-2000·sin(1000t)) = -20·sin(1000t) V.
Rewrite -20·sin using -sin(x) = cos(x + 90°):
v = 20·cos(1000t + 90°) V.
Plot the two phasors: current at angle 0 (magnitude 2), voltage at +90 (magnitude 20). Voltage is 90 degrees ahead of current — voltage leads, exactly ELI. Cross-check against the impedance: Z_L = jωL = j10 Ω, magnitude 10 at +90. Multiply the current phasor (2 at 0°) by Z (10 at +90°) and you get 20 at +90°, the same voltage phasor. The impedance angle is the phase lead.
Capacitor. Let C = 1 µF = 1×10⁻⁶ F, driven by v(t) = 5·cos(1000t) V. Then
i = C·dv/dt = 1×10⁻⁶ · d/dt[5·cos(1000t)] = 1×10⁻⁶ · (-5000·sin(1000t)) = -5×10⁻³·sin(1000t) A,
which is i = 5 mA · cos(1000t + 90°). Current at +90, voltage at 0 — current leads, exactly ICE. Cross-check: Z_C = -j/(1000 · 1×10⁻⁶) = -j·1000 Ω, magnitude 1000 at -90°. Divide the voltage phasor (5 at 0°) by Z (1000 at -90°) and you get 5 mA at +90°: current leads because dividing by a -90° impedance adds +90 to the angle.
The impedance angle and the derivative tell the same story. The +j in Z_L is the +90 the derivative gave the inductor's voltage; the -j in Z_C becomes +90 on the current once you divide. The j was never bookkeeping; it was the quarter-cycle shift wearing a different costume.
How to keep it straight
When the mnemonic blanks, don't guess. Write the constitutive law for the element you have. Inductor: v = L·di/dt. Capacitor: i = C·dv/dt. Whatever quantity sits under the derivative gets pushed +90 degrees, because differentiating a cosine adds 90 to its phase. Read off which one is ahead, and that's your lead. The reference confusion disappears too: you have two explicit phasors with explicit angles, not an abstract "ahead of what."
If you want the impedance as your check: a positive imaginary part (+j, the inductor) means voltage leads; a negative imaginary part (-j, the capacitor) means current leads.
This is the same habit that pays off elsewhere in circuits: regenerate a result from the defining relation instead of memorizing the outcome. It's why the small-signal gm tracks collector current rather than being a fixed number, and why the emitter voltage rises the moment the collector current does — both fall straight out of one equation the way the +90 falls out of the derivative here.
Check yourself
A capacitor is driven by v(t) = 3·cos(500t) V. Which statement about the current is correct?
A) Current lags voltage by 90 degrees. B) Current leads voltage by 90 degrees. C) Current is in phase with voltage. D) Voltage leads current by 90 degrees.
Correct answer: B.
Use i = C·dv/dt. The current is the derivative of the voltage, so the current is the quantity that gets pushed +90 degrees: i ∝ cos(500t + 90°), sitting ahead of the voltage at 0. Current leads voltage by 90 — that's ICE. B says exactly this. D ("voltage leads current") is the inductor's behavior (ELI), not the capacitor's; it's the inverted answer people land on when they flip ELI and ICE, so it is wrong here. A reverses the capacitor itself, and C forgets that a reactive element shifts the phase at all.
Close the gap
This sign keeps flipping because the mnemonic was never connected to anything you could rebuild. Once you've differentiated a cosine and watched the +90 appear, ELI and ICE stop being two strings to memorize and become one procedure you run. Gradual Learning is built to catch exactly this kind of gap — a rule you can recite but can't reconstruct — and to work it until you can regenerate it instead of recalling it.