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$C = Q/V$ from geometry: parallel plates, coaxial cylinders and spheres, dielectrics, and capacitors in series and parallel.
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By the end of this lesson you will be able to derive capacitances from Gauss's law, account for dielectrics, and combine capacitors.
From the last two lessons you can find fields with Gauss's law and potential differences by integrating the field. From Physics 2 you have met capacitors in circuits. This lesson computes capacitance from first principles for several shapes, and explains what dielectrics do and how capacitors combine.
| Term | What it means |
|---|---|
| Capacitor | Two conductors separated by an insulator, holding charges $+Q$ and $-Q$. |
| Capacitance | $C = Q/V$, charge stored per volt, in farads (C/V). |
| Farad | A very large unit; real capacitors are measured in μF, nF and pF. |
| Dielectric | An insulator between the plates; its molecules polarize and weaken the field. |
| Dielectric constant | $\kappa$, the factor by which a dielectric multiplies capacitance. |
| Parallel connection | Capacitors across the same voltage: $C = C_1 + C_2$. |
| Series connection | Capacitors carrying the same charge: $1/C = 1/C_1 + 1/C_2$. |
Put charge $+Q$ on one conductor and $-Q$ on another. The field between them, and so the potential difference $V$, is proportional to $Q$, because every field in electrostatics is. The ratio is the capacitance:
$$C = \frac{Q}{V}.$$
It depends only on the shapes, the spacing and the material between. To find it, suppose a charge $Q$, find the field with Gauss's law, integrate to get $V$, and divide. For parallel plates of area $A$ and separation $d$, $E = \sigma/\varepsilon_0 = Q/\varepsilon_0A$ and $V = Ed$, so
$$C = \frac{\varepsilon_0A}{d}.$$
Filling the gap with a dielectric multiplies $C$ by $\kappa$. Capacitors in parallel add, $C_1 + C_2$; in series, reciprocals add.
Another way: picture
Picture two metal plates facing each other, one covered in positive charge and one in negative, with a uniform field running straight across the gap. Pushing more charge on takes more work because the growing field pushes back. Bigger plates spread the charge thinner, a weaker field for the same charge; closer plates need less voltage for the same field. Both raise $C$.
Another way: steps
For plates much larger than their separation, the field between them is uniform, $E = \sigma/\varepsilon_0$, and zero outside, apart from a fringe at the edges. The potential difference is $V = Ed = Qd/\varepsilon_0A$, so $C = \varepsilon_0A/d$. With $\varepsilon_0 = 8.85$ pF/m, plates of $1$ m² a millimeter apart give only $8.85$ nF.
That is why a farad is enormous. Getting large capacitance takes huge area, tiny gaps, or a dielectric with large $\kappa$. Electrolytic capacitors roll up long strips of foil with an oxide layer nanometers thick; supercapacitors use porous carbon with thousands of square meters of surface per gram.
For coaxial cylinders of radii $a$ and $b$ and length $L$, the field between them is $\lambda/2\pi\varepsilon_0r$, and integrating from $a$ to $b$ gives $V = (\lambda/2\pi\varepsilon_0)\ln(b/a)$. So $C = 2\pi\varepsilon_0L/\ln(b/a)$, proportional to length, and depending on the radii only through their ratio.
For concentric spheres, the field is $kQ/r^2$ between them, so $V = kQ(1/a - 1/b)$ and $C = 4\pi\varepsilon_0ab/(b - a)$. Letting the outer sphere recede to infinity gives the capacitance of an isolated sphere, $4\pi\varepsilon_0R$. The Earth, treated this way, has a capacitance of only about $700$ μF.
Put an insulator between the plates and its molecules polarize: their positive charges shift slightly toward the negative plate and their negative charges toward the positive one. The layers of shifted charge at the surfaces of the dielectric partly cancel the plates' field, so for the same charge the voltage is smaller, by a factor $\kappa$, and the capacitance is larger by $\kappa$.
Air has $\kappa \approx 1.0006$, plastic films $2$ to $3$, glass about $5$, water $80$, and special ceramics thousands. A dielectric also resists breakdown, allowing stronger fields before it sparks through, which lets capacitors be both small and high in voltage.
Capacitors connected in parallel share the same voltage; the total charge is the sum, so $C = C_1 + C_2$. This is like increasing the plate area. Capacitors in series carry the same charge, since the plates joined between them start neutral and can only divide their charge; the voltages add, so $1/C = 1/C_1 + 1/C_2$. This is like increasing the gap.
A series combination is always smaller than its smallest member, and the smallest capacitor takes the largest share of the voltage. Engineers put capacitors in series to withstand a voltage higher than any one can alone.
What happens when you change a capacitor depends on whether it is still connected. Connected to a battery, $V$ is fixed: inserting a dielectric raises $C$ and draws more charge from the battery, $Q = \kappa CV$. Disconnected, $Q$ is fixed: inserting a dielectric lowers the voltage to $V/\kappa$.
Pulling the plates apart works the same way. With the battery connected, $C$ falls and charge flows back into the battery. Isolated, the charge stays, the field stays the same, and the voltage grows in proportion to the gap. Always decide first which of $Q$ and $V$ is held fixed.
Checking an answer. $Q$ must cancel from $C$. The units must reduce to $\varepsilon_0$ times a length: F/m times m. Moving conductors closer must raise $C$. For concentric cylinders or spheres with a small gap, the result must approach $\varepsilon_0A/d$ with $A$ the surface area and $d$ the gap, since a thin gap looks locally like flat plates.
Assuming charges $\pm Q$ is legitimate because the capacitance does not depend on them; any value gives the same ratio. Gauss's law gives the field only when the conductors have the symmetry of a sphere, cylinder or plane, which is why the textbook capacitors have those shapes. For other shapes, $C$ still exists but must be computed numerically.
The series rule depends on the connecting wire and inner plates being isolated from everything else, so that their net charge stays zero. If a third wire touched the junction, the capacitors would no longer be in series.
A modern phone contains hundreds of capacitors, most no bigger than a grain of sand, smoothing power supplies, filtering signals and storing charge briefly. Touchscreens sense your finger because it changes the capacitance between transparent electrodes. Camera flashes and defibrillators store charge and release it in a burst.
Memory chips store each bit as charge on a capacitor of a few tens of femtofarads. Keeping the capacitance up as cells shrink has driven decades of engineering: trenches etched deep into the silicon to increase area, and dielectrics with $\kappa$ of $20$ or more to replace silicon dioxide's $3.9$.
The parallel-plate formula assumes the field is uniform right to the edges and zero outside. Real plates have a fringing field that bulges out around the edges, adding a little capacitance. When the plates are wide compared with their gap, the fringe is a small fraction of the total and the formula is accurate to a percent or better.
For plates whose gap is comparable to their size, the fringe matters and the simple formula underestimates $C$. Engineers designing precision capacitors surround the plates with a guard ring held at the same potential, which pushes the fringe outside the measured region and makes $\varepsilon_0A/d$ exact for the central part. Such guard-ring capacitors were used at the National Bureau of Standards to measure $\varepsilon_0$ itself.
Capacitors combine in the opposite way from resistors, and seeing why fixes both rules. Resistors in series carry the same current and their voltages add, so resistances add. Capacitors in series carry the same charge and their voltages add, but a capacitor's voltage is $Q/C$, inversely proportional to $C$, so it is the reciprocals that add.
In parallel, resistors share a voltage and their currents add, so conductances, $1/R$, add. Capacitors share a voltage and their charges add, and a capacitor's charge is $CV$, proportional to $C$, so capacitances add directly. A good way to remember the difference: a capacitor's $C$ plays the role of a resistor's $1/R$, since each measures how much flows, charge or current, per volt.
This also explains why an equivalent capacitance can be found by the same step-by-step reduction used for resistor networks: find a group that is purely in series or purely in parallel, replace it by its equivalent, and repeat until one capacitor remains. Then work back outward, finding the charge and voltage on each original capacitor from the rules for the group it belonged to, exactly as in the network example above.
Each bit of the DRAM in a computer is a transistor and a capacitor of about $20$ to $30$ fF. Charged to about a volt, it holds on the order of a hundred thousand electrons for a 1, and none for a 0. The charge leaks away in milliseconds, so the chip reads and rewrites every cell many times a second, a process called refresh.
Memory makers, including Micron Technology in Boise, Idaho, have spent decades keeping the capacitance up as cells shrank to nanometers wide. They etch the capacitors as deep trenches or tall cylinders to gain area, and use dielectrics such as hafnium and zirconium oxides with $\kappa$ far above silicon dioxide's.
A phone's touchscreen is a grid of transparent electrodes, made of indium tin oxide, under the glass. Neighboring electrodes form tiny capacitors, and the controller measures their capacitance many times a second. A finger, which conducts, changes the field near the electrodes it approaches and so changes their capacitance by a fraction of a picofarad.
By locating which crossings changed, the controller finds each finger's position, which is how multi-touch works. Gloves defeat it because the insulating fabric keeps the finger too far away; touchscreen gloves weave conducting threads into the fingertips to restore the coupling.
Because $C = Q/V$, it can look as if more charge means more capacitance, or more voltage less. But $Q$ and $V$ rise together in proportion, and their ratio is fixed by the shape, spacing and dielectric. An uncharged capacitor has the same capacitance as a charged one, just as an empty bucket has the same volume as a full one.
A second error is to treat capacitors like resistors, adding them in series. Series capacitors combine like parallel resistors, and parallel capacitors add.
Plates of area $0.020$ m² are $0.50$ mm apart in air. Find the capacitance.
$C = \dfrac{8.85 \times 10^{-12} \times 0.020}{5.0 \times 10^{-4}} = 354\ \text{pF}$
$\varepsilon_0A/d$.
Connect it to a $12$ V battery. Find the charge.
$Q = CV = 354 \times 10^{-12} \times 12 = 4.2\ \text{nC}$
Charge per volt times volts.
Find the field between the plates.
$E = \dfrac{V}{d} = \dfrac{12}{5.0 \times 10^{-4}} = 2.4 \times 10^4\ \text{V/m}$
Uniform.
Slide in plastic with $\kappa = 2.5$, battery connected. Find the new charge.
$Q' = 2.5 \times 4.2 = 10.6\ \text{nC}$
$V$ fixed, $C$ up by $\kappa$.
Instead disconnect first, then insert it. Find the voltage.
$V' = \dfrac{12}{2.5} = 4.8\ \text{V}$
$Q$ fixed.
Concentric spheres have radii $10$ cm and $12$ cm. Find the field between them for charge $Q$.
$E = \dfrac{kQ}{r^2}$
Gauss's law with a sphere between them.
Integrate for the voltage.
$V = kQ\left(\dfrac{1}{0.10} - \dfrac{1}{0.12}\right) = kQ \times 1.667$
From inner to outer.
Divide for the capacitance.
$C = \dfrac{1}{k \times 1.667} = \dfrac{1}{9.0 \times 10^9 \times 1.667} = 66.7\ \text{pF}$
$Q$ cancels.
Compare with flat plates of the same area and gap.
$\dfrac{\varepsilon_0 \cdot 4\pi(0.11)^2}{0.020} = 67.3\ \text{pF}$
Close, since the gap is thin.
Remove the outer sphere to infinity.
$C = 4\pi\varepsilon_0 \times 0.10 = 11.1\ \text{pF}$
An isolated sphere.
Find the charge at $1000$ V on the isolated sphere.
$Q = 11.1 \times 10^{-12} \times 1000 = 11\ \text{nC}$
Much less than the pair holds.
A $2.0$ μF and a $4.0$ μF capacitor are in parallel, and the pair is in series with a $3.0$ μF capacitor across $18$ V. Combine the parallel pair.
$C_{24} = 2.0 + 4.0 = 6.0\ \mu\text{F}$
Same voltage across both.
Combine with the series capacitor.
$C = \dfrac{6.0 \times 3.0}{6.0 + 3.0} = 2.0\ \mu\text{F}$
Reciprocals add.
Find the total charge.
$Q = 2.0 \times 18 = 36\ \mu\text{C}$
On the series capacitor and on the pair.
Find the voltage across the $3.0$ μF capacitor.
$V_3 = \dfrac{36}{3.0} = 12\ \text{V}$
The smaller capacitance takes more voltage.
Find the voltage across the pair.
$V_{24} = \dfrac{36}{6.0} = 6.0\ \text{V}$
The two add to $18$ V.
Find the charge on the $2.0$ μF capacitor.
$Q_2 = 2.0 \times 6.0 = 12\ \mu\text{C}$
Parallel capacitors share the voltage.
Find the charge on the $4.0$ μF capacitor.
$Q_4 = 4.0 \times 6.0 = 24\ \mu\text{C}$
The two add to $36$ μC.
Combine the reciprocals.
$C = \dfrac{6.0 \times 12}{18} = 4.0\ \mu\text{F}$
Series.
Multiply by the voltage.
$Q = 4.0 \times 9.0$
The same charge on each.
Evaluate the charge.
A parallel-plate capacitor has capacitance $9$ pF. It is rebuilt with plates of twice the area, set half as far apart. What is its new capacitance?
Complete the worked solution: capacitors of $6$ μF and $3$ μF can be connected side by side or end to end across a $5$ V battery. Find the combined capacitance each way, in μF, and the charge on each capacitor when they are end to end, in μC.
Add the capacitances side by side.
$C_{\parallel} = C_1 + C_2 =$ p
Same voltage; the charges add.
Combine the reciprocals end to end.
$C_{\text{series}} = \dfrac{C_1C_2}{C_1 + C_2} =$ q
Same charge; the voltages add.
Multiply the series capacitance by the voltage.
$Q = C_{\text{series}}V =$ c
In series, each capacitor carries this same charge.
Check the voltages add up.
$\dfrac{Q}{C_1} + \dfrac{Q}{C_2} = V$
The smaller capacitor takes the larger voltage.
Match each capacitor or combination to its capacitance.
| $\varepsilon_0A/d$ | $4\pi\varepsilon_0R$ | $C_1 + C_2$ | $C_1C_2/(C_1 + C_2)$ | |
|---|---|---|---|---|
| parallel plates | ||||
| isolated sphere | ||||
| in parallel | ||||
| in series |
A $2$ μF capacitor is connected to a $5$ V battery. With the battery still connected, a slab of dielectric with $\kappa = 4$ is slid in to fill the gap. Fill in the charge before in μC, the capacitance after in μF, and the charge after in μC.
| value | |
|---|---|
| charge before (μC) | |
| capacitance after (μF) | |
| charge after (μC) |
An empty parallel-plate capacitor with square plates $10$ cm on a side has capacitance $70$ pF. A slab of dielectric with $\kappa = 3$, exactly as thick as the gap, is pushed in a distance $x$ cm from one edge. Write the capacitance, in pF, as a formula in $x$.
Answer:
A coaxial cable has an outer conductor whose radius is $3$ times that of the inner wire, with air between them. With $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m, what is its capacitance per meter, in pF/m?
Answer: pF per meter of cable
An engineer at a memory maker in Boise, Idaho, designs a DRAM cell whose capacitor is $24$ fF and is charged to $1.0$ V to store a 1. How many electrons does it hold, in thousands?
Answer: thousand electrons
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
An empty parallel-plate capacitor with square plates $10$ cm on a side has capacitance $20$ pF. A slab of dielectric with $\kappa = 4$, exactly as thick as the gap, is pushed in a distance $x$ cm from one edge. Write the capacitance, in pF, as a formula in $x$.
Answer:
You can find and combine capacitances. Explain to someone why two capacitors in series store less charge than either alone would at the same voltage.
21. Your turn: find the capacitance of $6.0$ μF and $12$ μF in series, and the charge on each across $9.0$ V., step 3
$Q = 36\ \mu\text{C}$
The $6.0$ μF one takes $6.0$ V, the other $3.0$ V.