Suppose you have a parallel plate capacitor that is disconnected from any power source and is discharged. At time $t=0$, the capacitor is connected to a battery. Make qualitative graphs of current ($I$) in the wire, charge ($Q$) on the positive plate, and voltage ($\Delta V$) across the capacitor over time.

• The capacitor is discharged.
• The capacitor is made up of parallel plates.
• The capacitor is connected to a power source at $t=0$.

• Graphs of $I$, $Q$, $\Delta V$$.

• The power source is connected correctly with respect to the capacitor and there are no other circuit elements (except for the wire).
• The wire itself has a small resistance, just so we do not have infinite current at $t=0$.
• Practically speaking, the capacitor becomes “fully charged” (with respect to the potential of the battery) at some finite time.

• We represent the setup below. The capacitor is pictured both disconnected and hooked up to the power source.

Before the capacitor is connected, we know that it is discharged, so there is a net neutral charge both on the wire and on the parallel plates. At time $t=0$, when the power source is connected, it brings with it an electric field, which the charges on the wire and capacitor do not immediately oppose, since they have not had time to accumulate. At time $t=0$ (or maybe slightly after 0, once the electric field has propagated at the speed of light), the electric field in the wire may look like this:

Slightly after time $t=0$, the current begins to exist and the capacitor begins to charge. Due to the high electric field at the very beginning, the charge flows very quickly at first. Since current is simply the movement of charge, our current graph will start at a very high value. However, we expect the current to drop pretty quickly, since the large flow of charge means a quickly charging capacitor, which begins to oppose the electric field and diminish it, slowing down the flow of charge. A few moments after time $t=0$, the circuit and electric field may look like this: