The two cases, with figures

The two possible signs correspond to the two orientations of the current relative to the enclosed plate.

Case 1: current entering the enclosed plate

The surface $\mathcal{S}$ contains the positive plate, and the conventional current in the wire points towards that plate. The charge inside $\mathcal{S}$ increases, so $dQ/dt > 0$; the incoming current is $+i$. Result:

$\ev{\frac{dQ}{dt} = +i}$

The current points towards the plate $+Q$ enclosed by $\mathcal{S}$: the charge grows, $dQ/dt = +i$.

Case 2: current leaving the enclosed plate

The surface $\mathcal{S}$ still contains the positive plate, but this time the conventional current $i$ is drawn leaving the plate. The charge inside $\mathcal{S}$ decreases: the current entering $\mathcal{S}$ is $-i$, so:

$\ev{\frac{dQ}{dt} = -i}$

The current moves away from the plate $+Q$: the charge falls, $dQ/dt = -i$.

Practical rule

If the arrow of $i$ points towards the plate labelled $+Q$, then $dQ/dt = +i$; if the arrow points away from it, then $dQ/dt = -i$.

Collegamenti

Argomenti: Electromagnetic induction Concetti: LC and RLC oscillations

Esercizi collegati: Problem — Natural frequency of an LC circuit · Problem — Capacitor of an AM radio receiver · Worked exercise — Forced RLC, find the generator’s EMF