Charging and Discharging a Capacitor

Edited by Gabriel Hood Spring 2026 Physics 2212)

A capacitor is a two-conductor device that stores energy in the electric field between its plates. When you connect an uncharged capacitor to a battery through a resistor, charge piles up on the plates over time — that's charging. Disconnect the battery and let the capacitor drive current through a resistor, and the stored charge bleeds back off — that's discharging. Both processes are governed by the same RC time constant [math]\displaystyle{ \tau = RC }[/math], and both follow exponential curves in time.

A capacitor is not a battery. A battery maintains a roughly constant emf from chemistry; a capacitor's voltage rises or falls as charge moves on or off its plates.

The Main Idea

A capacitor in a circuit acts like a charge reservoir. While charging, current flows from the battery and piles charge onto the plates; while discharging, the plates push current back through the resistor until they're empty. In both cases, the current at any instant is set by how much voltage the capacitor still has left to gain or lose, which is why both processes are exponential.

Charging

Charging a capacitor through a resistor (here, a light bulb). The field across the gap drops as charge accumulates, and current decays toward zero.

Connect an uncharged capacitor in series with a battery (emf [math]\displaystyle{ \mathcal{E} }[/math]) and a resistor. Initially there is no charge on the plates, so the full battery voltage drops across the resistor and the current is at its maximum, [math]\displaystyle{ I_0 = \mathcal{E}/R }[/math]. As charge builds up on the plates, it creates a back-voltage [math]\displaystyle{ V_C = Q/C }[/math] that opposes the battery. The net driving voltage [math]\displaystyle{ \mathcal{E} - Q/C }[/math] shrinks, and so does the current. Charging stops when [math]\displaystyle{ V_C = \mathcal{E} }[/math] and [math]\displaystyle{ Q = C\mathcal{E} }[/math].

In the bulb-and-capacitor demo, the bulb is brightest at [math]\displaystyle{ t = 0 }[/math] and dims to dark as [math]\displaystyle{ Q \to C\mathcal{E} }[/math].

Plate polarity. Conventional current flows out the + terminal of the battery. The plate it reaches first becomes positive (charge accumulates there); the other plate becomes negative. This determines the sign of [math]\displaystyle{ V_C }[/math] when applying Kirchhoff's loop rule.

Discharging

Discharging a charged capacitor through a resistor. The capacitor itself drives the current, which decays as Q drops.

Disconnect the battery and connect the charged capacitor across a resistor. Now the capacitor's voltage [math]\displaystyle{ V_C = Q/C }[/math] drives the current: [math]\displaystyle{ I = V_C/R = Q/(RC) }[/math]. As [math]\displaystyle{ Q }[/math] falls, so does the current — exponentially. The bulb starts bright and dims to dark, just like in charging, but for a different reason: now it's the capacitor running out of charge, not the capacitor opposing the battery.

Role of [math]\displaystyle{ R }[/math]. The resistance sets the timescale through [math]\displaystyle{ \tau = RC }[/math]. Larger [math]\displaystyle{ R }[/math] ⇒ slower charge/discharge. (A bulb with a thinner filament has higher [math]\displaystyle{ R }[/math], so it lights more dimly but for longer.) Smaller [math]\displaystyle{ R }[/math] means faster, brighter, and shorter.

A Mathematical Model

Parallel-plate capacitance. Treat the plates as two oppositely charged infinite sheets with surface charge density [math]\displaystyle{ \sigma = Q/A }[/math]. Each sheet contributes a field of magnitude [math]\displaystyle{ \sigma/(2\varepsilon_0) }[/math], and between the plates these fields add:

[math]\displaystyle{ E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A} }[/math]

The voltage across a gap of size [math]\displaystyle{ s }[/math] is [math]\displaystyle{ V = E\,s = Qs/(\varepsilon_0 A) }[/math], so

[math]\displaystyle{ C \equiv \frac{Q}{V} = \frac{\varepsilon_0 A}{s} }[/math]

Doubling the plate area doubles the capacitance; doubling the gap halves it.

Charging derivation. For a series RC circuit (battery emf [math]\displaystyle{ \mathcal{E} }[/math], resistor [math]\displaystyle{ R }[/math], initially uncharged capacitor [math]\displaystyle{ C }[/math]), Kirchhoff's loop rule gives

[math]\displaystyle{ \mathcal{E} - IR - \frac{Q}{C} = 0 }[/math]

Series RC circuit: battery emf ε, switch S, resistor R, capacitor C with charges +Q and −Q. Conventional current I = dQ/dt flows clockwise during charging.

Substituting [math]\displaystyle{ I = dQ/dt }[/math]:

[math]\displaystyle{ R\,\frac{dQ}{dt} = \mathcal{E} - \frac{Q}{C} = \frac{C\mathcal{E} - Q}{C} }[/math]

Separate variables and integrate from [math]\displaystyle{ (0, 0) }[/math] to [math]\displaystyle{ (t, Q) }[/math]:

[math]\displaystyle{ \int_0^Q \frac{dQ'}{C\mathcal{E} - Q'} = \int_0^t \frac{dt'}{RC} }[/math]

[math]\displaystyle{ -\ln\!\left(\frac{C\mathcal{E} - Q}{C\mathcal{E}}\right) = \frac{t}{RC} }[/math]

Solving for [math]\displaystyle{ Q }[/math]:

[math]\displaystyle{ Q(t) = C\mathcal{E}\left(1 - e^{-t/RC}\right) }[/math]

Differentiating gives the current:

[math]\displaystyle{ I(t) = \frac{dQ}{dt} = \frac{\mathcal{E}}{R}\,e^{-t/RC} }[/math]

Discharging derivation. Replace the battery with a wire. The loop rule becomes [math]\displaystyle{ IR + Q/C = 0 }[/math], i.e. [math]\displaystyle{ R\,dQ/dt = -Q/C }[/math], which integrates to

[math]\displaystyle{ Q(t) = Q_0\,e^{-t/RC}, \qquad I(t) = -\frac{dQ}{dt} = \frac{Q_0}{RC}\,e^{-t/RC} }[/math]

Charge on the capacitor vs. time. Charging (blue) reaches 63.2% of Q_max at t = τ; discharging (red) falls to 36.8% of Q₀ at t = τ. After 5τ both processes are essentially complete.

Time constant. The product [math]\displaystyle{ \tau = RC }[/math] has units of seconds and sets the timescale of every RC process. After one [math]\displaystyle{ \tau }[/math], a charging capacitor reaches [math]\displaystyle{ 1 - e^{-1} \approx 63.2\% }[/math] of its final charge; a discharging one falls to [math]\displaystyle{ e^{-1} \approx 36.8\% }[/math]. After [math]\displaystyle{ 5\tau }[/math] both are within 1% of their asymptote.

Diagrams (external): For a clean RC circuit schematic and the exponential charging/discharging curves, see Wikipedia: RC circuit (figures licensed CC BY-SA 4.0).

A Computational Model

The model below numerically integrates Kirchhoff's loop rule for the same series RC circuit derived above. Pick values for the battery emf [math]\displaystyle{ \mathcal{E} }[/math], resistance [math]\displaystyle{ R }[/math], and capacitance [math]\displaystyle{ C }[/math] at the top of the script and the simulation will plot [math]\displaystyle{ Q(t) }[/math] on the capacitor and [math]\displaystyle{ I(t) }[/math] through the resistor as the capacitor charges, then discharges. The numerical curves should match the analytic [math]\displaystyle{ Q(t) = C\mathcal{E}(1 - e^{-t/RC}) }[/math] and [math]\displaystyle{ Q(t) = Q_0\,e^{-t/RC} }[/math] from the section above.

Live simulation pending; sample output of charge buildup is shown below.

Sample output: Q(t) for a charging capacitor showing the characteristic exponential approach to Q_max = Cε.

Current and Charge within the Capacitors

The following graphs depict how current and charge within charging and discharging capacitors change over time.

When the capacitor begins to charge or discharge, current runs through the circuit. It follows logic that whether or not the capacitor is charging or discharging, when the plates begin to reach their equilibrium or zero, respectively, the current slows down to eventually become zero as well.

When the plates are charging or discharging, charge is either accumulating on either sides of the plates (against their natural attractions to the opposite charge) or moving towards the plate of opposite charge. While charging, until the electron current stops running at equilibrium, the charge on the plates will continue to increase until the point of equilibrium, at which point it levels off. Conversely, while discharging, the charge on the plates will continue to decrease until a charge of zero is reached.

Time Constant

The time constant of a circuit, with units of time, is the product of R and C. The time constant is the amount of time required for the charge on a charging capacitor to rise to 63% of its final value. The following are equations that result in a rough measure of how long it takes charge or current to reach equilibrium.

[math]\displaystyle{ Q = (C\mathcal{E})[1−e^{(−t/RC)}] }[/math]

[math]\displaystyle{ I = (\mathcal{E}/R)[e^{(−t/RC)}] }[/math]

• Note: [math]\displaystyle{ \mathcal{E} }[/math] is electromotive force(emf), whose units are Volts([math]\displaystyle{ V }[/math])

The Effect of Surface Area

For two different circuits, each with one of the above capacitors, the circuit with the second capacitor (with more surface area) has a current that stays more constant than the first. The larger capacitor also ends up with a greater amount of charge on its plates.

This is because fringe field magnitude is inversely proportional to plate area, as shown in the equation below. In the first, short time interval, roughly equal quantities of charge will accumulate on the capacitor plates. However, due to its greater area, capacitor 2 will have a weaker fringe field. This, in turn, results in a greater net field for that circuit. This greater net field results in more charge for that circuit compared to the other. More charge will be driven from the negative to the positive plate, and the drift speed changes less for capacitor 2 than capacitor 1.

The equation for fringe electric field is the following:

Examples

Simple

Question 1. For a parallel-plate capacitor with capacitance [math]\displaystyle{ C = \varepsilon_0 A / s }[/math], predict how each change affects [math]\displaystyle{ C }[/math] and the field [math]\displaystyle{ E }[/math] between the plates (the plates carry a fixed charge [math]\displaystyle{ Q }[/math]).

(a) Doubling the plate radius:

A) quarters [math]\displaystyle{ C }[/math] B) halves [math]\displaystyle{ C }[/math] C) doubles [math]\displaystyle{ C }[/math] D) quadruples [math]\displaystyle{ C }[/math]

(b) Doubling the plate radius (effect on [math]\displaystyle{ E }[/math] with fixed [math]\displaystyle{ Q }[/math]):

A) quarters [math]\displaystyle{ E }[/math] B) halves [math]\displaystyle{ E }[/math] C) doubles [math]\displaystyle{ E }[/math] D) quadruples [math]\displaystyle{ E }[/math]

(c) Doubling the plate separation [math]\displaystyle{ s }[/math]:

A) quarters [math]\displaystyle{ C }[/math] B) halves [math]\displaystyle{ C }[/math] C) doubles [math]\displaystyle{ C }[/math] D) quadruples [math]\displaystyle{ C }[/math]

(d) Doubling the capacitance (with [math]\displaystyle{ Q }[/math] fixed):

A) quarters [math]\displaystyle{ E }[/math] B) halves [math]\displaystyle{ E }[/math] C) doubles [math]\displaystyle{ E }[/math] D) quadruples [math]\displaystyle{ E }[/math]

Answers: (a) D, (b) A, (c) B, (d) B.

Why:

• (a) Plate area is [math]\displaystyle{ A = \pi r^2 }[/math]. Doubling [math]\displaystyle{ r }[/math] gives [math]\displaystyle{ A \to 4A }[/math], and since [math]\displaystyle{ C = \varepsilon_0 A/s }[/math], we get [math]\displaystyle{ C \to 4C }[/math].
• (b) The field is [math]\displaystyle{ E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A) }[/math]. With [math]\displaystyle{ Q }[/math] fixed and [math]\displaystyle{ A \to 4A }[/math], [math]\displaystyle{ E \to E/4 }[/math].
• (c) [math]\displaystyle{ C \propto 1/s }[/math], so doubling [math]\displaystyle{ s }[/math] halves [math]\displaystyle{ C }[/math].
• (d) With [math]\displaystyle{ Q }[/math] fixed, doubling [math]\displaystyle{ C }[/math] halves [math]\displaystyle{ V = Q/C }[/math], and since [math]\displaystyle{ E = V/s }[/math], the field also halves.

Middling

Question 2. A battery is connected through a switch to a resistor (light bulb) in series with an initially uncharged capacitor.

Series circuit: battery, switch, bulb (resistor), capacitor.

What is the current at points A, B, and C (a) the instant the switch closes, and (b) after a long time? What is the final charge on the capacitor?

Solution. At [math]\displaystyle{ t=0 }[/math] the capacitor is uncharged, so [math]\displaystyle{ V_C = 0 }[/math] and the loop equation gives [math]\displaystyle{ I_0 = \varepsilon/R }[/math]. Because the circuit is a single series loop, the current is the same at A, B, and C: all equal to [math]\displaystyle{ \varepsilon/R }[/math].

As [math]\displaystyle{ t \to \infty }[/math], the capacitor charges until [math]\displaystyle{ V_C = \varepsilon }[/math]. With no voltage across [math]\displaystyle{ R }[/math], the current at every point is zero. The final charge is [math]\displaystyle{ Q_\infty = C\varepsilon }[/math].

Worked answers for Question 2.

Difficult

Question 3. The switch in the circuit below has been closed for a long time, then is opened.

Circuit with a battery, capacitor, and bulb; analyze before and after opening the switch.

(a) Before the switch opens (steady state): What is the current at each point? What is the charge on the capacitor? Is the bulb lit?

(b) Immediately after the switch opens: Is the bulb lit? After a long time? What is the initial current through the bulb, and in what direction?

Solution.

(a) In steady state the capacitor is fully charged and acts as an open circuit, so no current flows in the branch containing the capacitor. Current still flows through any branch that bypasses the capacitor (for example the bulb in parallel with the source), set by Ohm's law on that loop. The capacitor charge is [math]\displaystyle{ Q = C V_C }[/math], where [math]\displaystyle{ V_C }[/math] equals the steady-state voltage across its terminals.

(b) When the switch opens, the battery is disconnected. The charged capacitor now acts as the only EMF source and discharges through the bulb. Immediately after, the bulb is lit with current [math]\displaystyle{ I_0 = V_C/R }[/math], where [math]\displaystyle{ V_C }[/math] is the capacitor voltage just before opening. The current direction reverses relative to the charging direction because the capacitor is now driving the loop. The current decays as [math]\displaystyle{ I(t) = I_0 e^{-t/RC} }[/math], so after several time constants the bulb goes dark.

Worked answers for Question 3.

Connectedness

Capacitor can be temporary batteries. Capacitors in parallel can continue to supply current to the circuit if the battery runs out. This is interesting because the capacitor gets its charge from being connected to a chemical battery, but the capacitor itself supplies voltage without chemicals.

Capacitors are being researched for applications in electromagnetic armour and electromagnetic weapons. Currently capacitors are used as detonators in nuclear weapons. Capacitors also are largely involved in separations of AC and DC components.

History

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why. In 1745 Ewald Georg von Kleist was the first to "discover" capacitors in Germany. He connected a generator to glass jars of water and charged them. When he touched the wire they were connected to he shocked himself (discharged the capacitor). At the same time Pieter van Musschenbroek made a similar capacitor and named it the Leyden Jar. When Benjamin Franklin studied the Leyden Jar he determined, among other things, that the charge was stored on the glass. During his studies Franklin was the first to give the capacitor the name battery. Since then batteries have most often been electro-chemical cells of capacitors made of sheets of conducting and dielectric material.

See also

Further reading

1. Williams, Henry Smith. "A History of Science Volume II, Part VI: The Leyden Jar Discovered"
2. Keithley, Joseph F. (1999). The Story of Electrical and Magnetic Measurements: From 500 BC to the 1940s

External links

1. Wikipedia Page "Capacitor"[1]
2. Khan Academy[2]
3. WebAssign "Lab 4 - Charge and Discharge of a Capacitor"[3]
4. "Charge of Capacitor vs Current" [4]

References

1. Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions. 3rd ed. Vol. 2. N.p.: John Wiley and Sons, 2002. Print.
2. "Capacitor." Wikipedia. Wikimedia Foundation, n.d. Web. 05 Dec. 2015.