Source: Physics 212 Lab 4 Activity v1.2 | Course: University Physics: Electricity & Magnetism
Tags: capacitors, charge conservation, parallel capacitors, voltage, Q=CV, electrolytic capacitor, DAC, IOLab, charge redistribution
Difficulty: Introductory-Intermediate | Prerequisites: Basic understanding of voltage, current, and simple DC circuits. Familiarity with Prelab 4 (charging a capacitor with the IOLab DAC) is helpful.
This lab sits at the heart of electrostatics in PHYS 212: you already know a capacitor stores charge, and now you test whether that charge is genuinely conserved when it redistributes across multiple capacitors. The experiment bridges the gap between the textbook formula Q = CV and a hands-on, quantitative measurement using the IOLab. If you missed Prelab 4, the key background is that you charged a single capacitor via the IOLab's DAC output, measured its voltage, then observed what happened when you disconnected the voltage source. Understanding this lab well prepares you for series and parallel capacitor networks in upcoming problem sets and exams.
When you charge a capacitor and then disconnect the voltage source, the charge on that capacitor stays put. If you then connect uncharged capacitors in parallel, the charge redistributes so the voltage drops, but the total charge across all capacitors remains the same. You test this by comparing Q_total before and after adding capacitors in parallel, using Q = CV and IOLab voltage measurements.
Capacitor (C)
A two-terminal device that stores electrical energy by holding equal and opposite charges on its plates. In simple terms, think of it as a tiny rechargeable charge bucket.
Capacitance
The ratio of charge stored to voltage applied, measured in farads (F). C = Q / V. In simple terms, it tells you how much charge a capacitor can hold per volt.
Electrolytic capacitor
A polarised capacitor with a marked positive (+) terminal that must be connected in the correct orientation. Commonly used in labs because they offer large capacitance in a small package. Get the polarity wrong and you can damage it.
DAC (digital-to-analogue converter)
The IOLab's programmable voltage output. In this lab it acts as your voltage source, supplying a known, steady voltage to charge the capacitor.
GND (ground)
The IOLab's reference point at 0 V. The negative side of your voltage source in these circuits.
Charge conservation
The principle that the total electric charge in an isolated system remains constant over time. Charge can move between components, but it cannot appear from or vanish into nothing.
Parallel connection
Components wired so they share the same two nodes (same voltage across each). When capacitors are in parallel, their capacitances add: C_total = C_1 + C_2 + ... + C_n.
V_1 and V_2
Measurement points on the circuit where the IOLab's analogue inputs read voltage relative to IOLab ground. V_1 sits at the top of the capacitor, V_2 at the bottom. The voltage across the capacitor is V_1 minus V_2.
Charge redistribution
When a charged capacitor is connected in parallel with uncharged capacitors (DAC disconnected), charge flows from the charged capacitor to the others until all share the same voltage. The total charge does not change.
Circuit 1 (Figure 1a): the capacitor is wired between the DAC (positive) and GND (negative). The IOLab measures voltage at V_1 (top plate) and V_2 (bottom plate).
When you set the DAC to, say, 1 V, charge flows onto the capacitor plates until the voltage across the capacitor matches the DAC output.
The capacitor is fully charged when V_1 minus V_2 equals the DAC voltage and current has effectively stopped.
Circuit 2 (Figure 1b): the wire from DAC to the capacitor's positive plate is physically removed. GND remains connected.
With the DAC disconnected, no path exists for charge to enter or leave the capacitor through the DAC.
The voltage across the capacitor should remain at its charged value (you observed this in Prelab 4). Small drift may occur from leakage, but the charge is essentially trapped.
Start with one charged capacitor C_1 (DAC already disconnected). Record its voltage.
While keeping the DAC wire disconnected and GND still connected, plug one or more uncharged capacitors (C_2, C_3, ...) in parallel with C_1 on the breadboard.
All capacitors now share the same two nodes, so they must settle to the same voltage.
Because no external source is connected, the only charge available is what was already on C_1. That charge spreads across all the capacitors.
Before adding capacitors: Q_initial = C_1 * V_initial (only C_1 holds charge).
After adding capacitors: all capacitors share a new, lower voltage V_final. Total charge is Q_final = (C_1 + C_2 + ... + C_n) * V_final.
The claim under test: Q_initial = Q_final. If this holds (within measurement uncertainty), charge is conserved.
Adding uncharged capacitors in parallel increases the total capacitance.
For the same total charge, a larger capacitance means a smaller voltage (Q = CV rearranges to V = Q / C_total).
The voltage drop does not mean charge was lost. It means the same charge is now spread across a bigger "container."
Charge on a capacitor: Q = C * V where Q is charge in coulombs, C is capacitance in farads, and V is voltage in volts.
Voltage across the capacitor from IOLab measurements: V_cap = V_1 - V_2 where V_1 and V_2 are the analogue input readings at the top and bottom of the capacitor.
Total capacitance for capacitors in parallel: C_total = C_1 + C_2 + C_3 + ... + C_n Capacitances simply add.
Charge conservation test: Q_before = C_1 * V_initial Q_after = (C_1 + C_2 + ... + C_n) * V_final If Q_before and Q_after agree within experimental uncertainty, the claim is supported.
Uncertainty propagation (simplified): If the main uncertainty is in voltage, then the uncertainty in Q is roughly delta_Q = C * delta_V, where delta_V is the uncertainty in your voltage measurement. When comparing Q_before and Q_after, check whether their ranges overlap.
Charge conservation is the reason camera flashes work: a capacitor charges slowly from a battery, then dumps all its stored charge through the flash bulb in milliseconds. The charge that went in equals the charge that comes out. The same principle governs how defibrillators deliver a precise jolt to a patient's heart, and why engineers must carefully account for charge sharing when designing memory circuits in microprocessors.
"The voltage stays the same when I add capacitors in parallel." It does not. The voltage drops because the same charge is now spread across a larger total capacitance. The charge is conserved, not the voltage.
"Charge leaks away the moment I disconnect the DAC." In an ideal circuit, charge stays on the capacitor indefinitely. In practice there is slow leakage, but over the timescale of this experiment it is negligible.
"Capacitors in parallel all have different voltages." By definition, parallel components share the same voltage across their terminals.
"Adding more capacitors in parallel increases the total charge." No external source is connected, so no new charge enters the system. The total charge is fixed; only the distribution changes.
⚠️ You will almost certainly be asked to calculate total charge before and after connecting capacitors in parallel. Make sure you can do Q = CV quickly and accurately.
⚠️ Exam questions often ask: "Does voltage change when capacitors are connected in parallel with a disconnected source?" Yes. "Does total charge change?" No.
⚠️ Know the difference between what is conserved (charge, when the system is isolated) and what is not (voltage across individual capacitors, energy stored).
⚠️ Energy is not conserved in this process. Some energy is lost as heat in the wires during charge redistribution. This is a classic follow-up exam question.
⚠️ Uncertainty analysis: you need to show that Q_before and Q_after agree within experimental error to support the claim. Overlapping uncertainty ranges = claim supported.
True or False: When capacitors are connected in parallel, the voltage across each capacitor is different. (False, they share the same voltage.)
True or False: Disconnecting the DAC removes all charge from the capacitor. (False, the charge stays on the plates.)
Fill in the blank: The total capacitance of three 100 µF capacitors in parallel is ______ µF. (300 µF)
True or False: If you add uncharged capacitors in parallel to a charged one (DAC disconnected), the total charge increases. (False, total charge is conserved.)
Fill in the blank: If Q_initial = 100 µC on a single capacitor and you add two identical uncharged capacitors in parallel, the final voltage across each is ______ of the original voltage. (One third)
Q: A 220 µF capacitor is charged to 1.0 V, then the voltage source is disconnected. A second, uncharged 220 µF capacitor is connected in parallel. What is the final voltage across both capacitors?
A: Total charge is Q = 220 µF × 1.0 V = 220 µC. Total capacitance is now 440 µF. V_final = Q / C_total = 220 µC / 440 µF = 0.50 V.
Q: In the scenario above, what is the charge on each capacitor after they reach equilibrium?
A: Both capacitors sit at 0.50 V. Each has Q = 220 µF × 0.50 V = 110 µC. Total = 110 + 110 = 220 µC, matching Q_initial.
Q: You measure V_initial = 0.95 V ± 0.02 V on a 100 µF capacitor (DAC disconnected). You then add a 47 µF capacitor in parallel and measure V_final = 0.64 V ± 0.02 V. Does the data support charge conservation?
A: Q_before = 100 µF × 0.95 V = 95 µC (uncertainty: 100 × 0.02 = ±2 µC, so 93 to 97 µC). Q_after = (100 + 47) µF × 0.64 V = 94.1 µC (uncertainty: 147 × 0.02 = ±2.9 µC, so 91.2 to 97.0 µC). The ranges overlap, so the data is consistent with charge conservation.
Q: Why does the total energy stored in the capacitors decrease after charge redistribution, even though charge is conserved?
A: Energy stored is U = ½CV². When charge redistributes to a lower voltage across a larger capacitance, the total energy decreases. The "missing" energy was dissipated as heat in the resistance of the connecting wires during the brief current surge.
Q: If you forgot to disconnect the DAC before adding a second capacitor, would your experiment still test charge conservation? Why or why not?
A: No. With the DAC connected, it would supply additional charge to maintain its set voltage. The system would no longer be isolated, so you could not attribute any change in total charge (or lack thereof) purely to conservation.
This connects directly to series vs. parallel capacitor networks (upcoming in PHYS 212). In parallel, capacitances add and charge splits; in series, the reciprocals add and charge on each capacitor is the same. The energy loss during charge redistribution links to resistive dissipation, which you will study more carefully in RC circuits. Charge conservation itself is a universal principle that reappears in Kirchhoff's junction rule for current: the total current entering a node equals the total current leaving, because charge is neither created nor destroyed.
capacitor, capacitance, charge conservation, Q = CV, parallel capacitors, charge sharing, charge redistribution, voltage drop, IOLab, DAC, GND, breadboard, electrolytic capacitor, farads, microfarads, µF, RC circuit, energy dissipation, Kirchhoff's junction rule, PHYS 212, Physics 212, University of Illinois, claim testing, uncertainty analysis, isolated system