Which quantity is held fixed: a connected capacitor behaves nothing like an isolated one

01The mistake

A parallel-plate capacitor is charged and the plate separation is then doubled. Ask what happens to the stored energy. There is no single answer — it depends on whether the battery is still connected — and most students give one anyway. With the battery attached the energy halves; isolated, it doubles.

The dielectric version is the same trap with a different variable. Insert a slab with the battery connected and the charge rises, the field between the plates is unchanged, and the stored energy rises. Insert it after disconnecting and the charge is fixed, the field drops, the voltage drops, and the stored energy falls. Opposite directions for the energy from the same physical action.

The deeper tell is a student who treats $Q$ and $V$ as both fixed. Asked what changes when $C$ changes, they answer that nothing does, since $Q = CV$ has to hold — which it does, and exactly one of the two can be pinned at a time. Naming which one is the entire setup step.

Students also pick the wrong energy formula, which is the same error wearing different clothes. $U = \frac{1}{2}CV^2$ is convenient at constant $V$ and $U = Q^2/2C$ at constant $Q$. Both are always true; choosing the one whose fixed quantity matches the constraint is what makes the answer readable in a line instead of three.

02Why it makes sense to the student

The problems look identical. Two plates, a slab or a separation change, and the only difference is one clause about a switch or a disconnected wire. That clause carries the entire physical content and it occupies about five words of a long stem.

Capacitor problems are usually taught as a chain of consequences, and students memorize the chain. Increase the separation, decrease the capacitance, increase the voltage, and so on — a sequence that is correct for exactly one of the two cases and gets reproduced for both.

The battery's role is passive and therefore easy to forget. A battery that is simply sitting there connected feels like background, not like a constraint that is actively moving charge to keep the voltage fixed. Students do not picture charge flowing from the battery during the plate separation, though that is precisely what happens.

And the energy bookkeeping has a second term that is easy to miss. With the battery connected, the work done by the person pulling the plates is not the only energy flow; the battery gains or loses energy too. Students trying to reconcile their answer with energy conservation often conclude they must have made an arithmetic error when they have actually found a real effect.

03The correction

Make the first line of every capacitor solution a declaration: battery connected, so $V$ is constant; or isolated, so $Q$ is constant. Write it down before touching a formula. Most of the error disappears at that step because it forces the student to find the clause in the stem.

Then work the chain in the right order from the pinned quantity. Constant $V$: compute the new $C$, get $Q = CV$, get $U = \frac{1}{2}CV^2$. Constant $Q$: compute the new $C$, get $V = Q/C$, get $U = Q^2/2C$. Two chains, each three steps, and the choice of chain is the whole problem.

Work the same physical change both ways, side by side on the board, and circle the quantities that moved in opposite directions. Doubling the separation halves the stored energy with a battery and doubles it without. Seeing one action produce opposite results makes the constraint feel causal rather than bureaucratic.

Give the mechanism for the connected case out loud: as the plates separate with the battery attached, the capacitance falls, so charge flows back through the battery to keep $V$ fixed. Students who can describe charge actually moving stop treating the battery as scenery.

Then close the energy accounting for the connected case, since this is where careful students get stuck. The work done pulling the plates apart plus the energy returned to the battery accounts for the change in stored energy. The stored energy decreasing while the person does positive work is not a contradiction once the battery's term is in the ledger.

04A sample question

05What each wrong answer reveals

• A The isolated chain applied to a connected capacitor. The dominant wrong answer, and it is the correct answer to the other problem. Ask what the battery is doing during the separation. Once the student says it holds the voltage at $V_0$, $Q = CV$ with a halved $C$ gives them the charge, and the rest follows.
• B Correct. $C$ halves at constant $V_0$, so $Q = CV_0$ halves and $U = \frac{1}{2}CV_0^2$ halves with it.
• C Charge handled correctly, energy left alone. This student applied the constant-voltage constraint properly to get the charge and then reasoned about energy from $V$ alone. Both $C$ and $V$ appear in $U = \frac{1}{2}CV^2$, and only one of them was held fixed. Half-right in a recoverable way: the constraint is understood and one formula was read incompletely.
• D Right formula, wrong constraint. $U = Q^2/2C$ is always true, and this student chose the form suited to a constant-charge problem. Since $Q$ is not constant here, holding it fixed in that formula produces the wrong answer by the same factor as the charge error. Worth noting that they land on a halved energy, which is the correct result, reached by two compensating mistakes — the kind of agreement that makes a misconception very hard to detect from answers alone.

A has the wrong constraint throughout. C has the right constraint and an incomplete formula. D reaches the correct energy from the wrong constraint, which is the most dangerous of the three, since nothing in the final number signals the error. Always read which quantity a student declared fixed, not only what they computed.

06Try it in Mistake Master