A complete guide to the parallel plate capacitor, from how it stores charge to how much energy it holds.
Introduction
Open almost any electronic device, from a phone charger to a camera flash to a power supply, and you will find capacitors. They are small components with a simple job: store electric charge and, with it, electrical energy. In this post we build up the whole picture step by step:
- What a capacitor is and how it works
- The capacitance of a parallel plate capacitor
- What happens when a dielectric is inserted, and the dielectric constant
- The electric field in the different regions of the capacitor
- The energy stored in a capacitor
1. What is a capacitor?
A capacitor is a device that stores electric charge. In its simplest form, the parallel plate capacitor, it consists of two conducting plates separated by a small gap. The gap is filled with vacuum, air, or an insulating material called a dielectric.
The first picture in our set shows both sides of the idea: on the left, the idealised model of two parallel metal plates, each with a lead attached; on the right, a real-world electrolytic capacitor, a cylindrical component with two wire leads and the ratings printed on its body (here 47 µF, 450 V). The cylinder looks very different from two flat plates, but inside it the same principle applies: two conductors, separated by an insulator, with a large surface area folded or rolled into a compact shape.
How a capacitor works
When the two plates are connected to a battery:
- The battery pulls electrons from one plate and pushes them onto the other.
- One plate becomes positively charged (+Q) and the other negatively charged (−Q).
- An electric field develops in the gap between the plates, pointing from the positive plate to the negative plate.
- The potential difference between the plates rises until it equals the battery voltage, and the flow of charge stops.
The two plates always carry equal and opposite charges, and the charge stored is directly proportional to the potential difference across the plates:
Q = CV
The constant of proportionality C is the capacitance. It is measured in farads (F); one farad is one coulomb per volt. In practice we use microfarads (µF), nanofarads (nF) and picofarads (pF), since a farad is a very large unit.
2. Capacitance of a parallel plate capacitor (vacuum between the plates)
For two parallel plates with vacuum between them, the capacitance depends only on geometry:
C₀ = ε₀A / d
where:
- A is the area of each plate
- d is the separation between the plates
- ε₀ is the permittivity of free space (about 8.85 × 10⁻¹² F/m)
This formula tells us how to build a capacitor with large capacitance:
- Increase the plate area A. More area gives the charge more room to spread out, so more charge can be stored per volt. This is why real capacitors use long rolled-up foils.
- Decrease the separation d. Plates that are closer together produce a stronger attraction between the opposite charges, which lets more charge sit on the plates at the same voltage.
- Use a dielectric between the plates, which is the next topic.
Note that capacitance is a property of the capacitor’s geometry and the material between the plates. It does not depend on the charge or the voltage; those two always adjust so that Q/V stays constant.
3. Capacitance with a dielectric
A dielectric is an insulating material, such as paper, glass, mica, ceramic or plastic. When it is placed between the plates, the capacitance goes up. The factor by which it increases is called the dielectric constant (K), also written κ or the relative permittivity:
K = C / C₀
In words: K is the capacitance with the dielectric between the plates divided by the capacitance with vacuum between the plates. Since K is a ratio of two capacitances, it has no units. For vacuum K = 1, and for every material it is greater than 1.
For a capacitor completely filled with the dielectric:
C = K C₀ = Kε₀A / d
Why does a dielectric increase the capacitance?
The answer lies in polarisation, and the second diagram in our set illustrates it well.
In the diagram, the top plate carries a surface charge density +σ and the bottom plate −σ. A dielectric slab of thickness t sits in the gap of total width d. The free charges on the plates create an electric field, labelled E_free, pointing downward from the positive plate to the negative one.
This field acts on the atoms and molecules of the dielectric. Their positive and negative charges are pulled slightly in opposite directions, so the slab becomes polarised:
- The face of the dielectric nearest the positive plate acquires a layer of negative induced (bound) charge, −σ_ind.
- The face nearest the negative plate acquires a layer of positive induced charge, +σ_ind.
These induced surface charges are not free to move; they are bound to the atoms. But they still create their own field, E_ind, and this field points upward, opposite to E_free.
Inside the dielectric the two fields partly cancel, so the net field is reduced:
E_net = E_free − E_ind
With a weaker field in the gap, less work is needed to move a unit charge from one plate to the other, which means the potential difference across the plates drops for the same stored charge. Since C = Q/V, a smaller V for the same Q means a larger capacitance.
4. The electric field in the inner and outer regions of the capacitor
To understand the field properly, we need to distinguish the regions in and around the capacitor.
Outside the plates
For an ideal parallel plate capacitor (plates large compared with their separation), the fields of the two oppositely charged plates cancel outside the plates. The field in the outer region is essentially zero. All the field is confined to the gap between the plates. (Near the edges the field lines bulge outward, called fringing, which we neglect in the ideal case.)
Between the plates, without a dielectric
The field is uniform and given by:
E₀ = σ / ε₀ = Q / (ε₀A)
and the potential difference is V₀ = E₀d.
Between the plates, in the gap regions not occupied by the dielectric
In the diagram, the air gaps above and below the slab carry the field E_free, which is the field due to the free charge on the plates alone. This equals σ/ε₀, the same as the field without any dielectric.
Inside the dielectric (the inner region)
Here the net field is reduced by the dielectric constant:
E_net = E_free − E_ind = E_free / K
Rearranged, this gives another way to see what the dielectric constant means:
K = E₀ / E = E_free / E_net
That is, K is the field without the dielectric divided by the net field with the dielectric. The induced field is therefore E_ind = E_free (1 − 1/K).
Putting it together for a partly filled gap
If the dielectric has thickness t in a gap of d, the potential difference is the sum over the regions:
V = E_free (d − t) + E_free t / K
So the capacitance becomes:
C = ε₀A / [ (d − t) + t/K ]
Check the limits: if t = d (fully filled) you get C = Kε₀A/d, and if t = 0 (no dielectric) you get back C₀ = ε₀A/d. Both agree with what we found above.
5. Energy stored in a capacitor
Charging a capacitor takes work, and that work is stored as electrical potential energy.
The circuit
The circuit diagram in our set shows how this can be studied in the lab. It contains a battery (Bt), a key (K), a rheostat, an ammeter, the capacitor (C), and a voltmeter connected across the capacitor. Closing the key lets the battery charge the capacitor; the rheostat controls the charging current, and the voltmeter reads the voltage V as the charge Q builds up.
Deriving the energy
Suppose the capacitor already holds charge q, so its voltage is v = q/C. To move a small extra charge dq from one plate to the other, the work required is:
dW = v dq = (q / C) dq
Adding up the work as the charge grows from 0 to Q:
W = ∫₀^Q (q / C) dq = Q² / 2C
The graphical view
The graph in our set plots Q against V. Because Q = CV, this is a straight line through the origin whose slope is the capacitance C. The work done in charging is the area under the line, which is the red triangle labelled E:
Area = ½ × base × height = ½ × V × Q
The formulas
This work is stored as the electrical potential energy U of the capacitor. Using Q = CV, it can be written in three equivalent forms:
U = Q² / 2C = ½ CV² = ½ QV
Which one to use depends on what you know:
- Know Q and C? Use Q²/2C.
- Know C and V? Use ½CV².
- Know Q and V? Use ½QV.
Worked example
The capacitor in the first picture is rated 47 µF and 450 V. Fully charged to its rated voltage, the stored energy would be:
U = ½ × (47 × 10⁻⁶ F) × (450 V)² ≈ 4.76 J
That may sound small, but it can be released in a fraction of a millisecond, which is why such capacitors are used in camera flashes and power supplies, and why a charged high-voltage capacitor must be handled with real care.
Where is the energy stored?
The energy sits in the electric field between the plates. Introducing a dielectric changes this: at a fixed voltage, the capacitor stores K times more energy; at a fixed charge, it stores K times less energy. In the latter case the missing energy is the work done in pulling the dielectric into the gap, so the slab is drawn in by the field.
Quick summary
| Quantity | Formula |
|---|---|
| Charge on a capacitor | Q = CV |
| Capacitance (vacuum) | C₀ = ε₀A / d |
| Dielectric constant | K = C / C₀ = E₀ / E = E_free / E_net |
| Capacitance with dielectric (full) | C = Kε₀A / d |
| Net field inside the dielectric | E_net = E_free − E_ind = E_free / K |
| Capacitance with a slab of thickness t | C = ε₀A / [(d − t) + t/K] |
| Energy stored | U = Q²/2C = ½CV² = ½QV |
| U = Q²/2C = ½CV² = ½QV |
Key takeaways
- A capacitor stores charge on two conductors separated by an insulator, and its capacitance C = Q/V depends only on its geometry and the material in the gap.
- Capacitance grows with plate area and shrinks with plate separation.
- A dielectric polarises, creates an opposing field E_ind, and weakens the net field inside it to E_free/K, which raises the capacitance by the factor K.
- Outside the plates the ideal field is zero, and between the plates it is uniform.
- The energy stored equals the area under the Q–V graph: ½QV = ½CV² = Q²/2C.
Worked examples
The six examples below follow the order of the post. Examples 2 to 6 reuse one capacitor, so you can see how the quantities connect.
Example 1: Charge stored (Q = CV)
A 100 µF capacitor is connected to a 12 V battery. How much charge does it store?
Example 2: Capacitance of a parallel plate capacitor
Two square plates of area A = 0.02 m² (200 cm²) are separated by d = 1 mm of air (treat air as vacuum). Find C₀.
Halving the gap to 0.5 mm would double this to 354 pF; halving the plate area would halve it to 88.5 pF.
Example 3: Capacitance with a dielectric
The gap of the capacitor in Example 2 is completely filled with mica, which has K = 6. Find the new capacitance.
If it is connected to a 100 V battery, the charge rises from Q₀ = C₀V = 17.7 nC without the mica to Q = CV = 106.2 nC with it.
Example 4: Electric field inside and outside the dielectric
The capacitor from Example 2 is charged to 100 V, then disconnected from the battery, so its charge stays fixed at 17.7 nC. Mica (K = 6) is then slipped in to fill the gap. Find the fields and the new voltage.
Field before the mica, which is also E_free because the plate charge has not changed:
The capacitor from Example 2 is charged to 100 V, then disconnected from the battery, so its charge stays fixed at 17.7 nC. Mica (K = 6) is then slipped in to fill the gap. Find the fields and the new voltage.
Field before the mica, which is also E_free because the plate charge has not changed:
Net field inside the mica, and the induced field that cancels part of E_free:
New voltage across the plates:
The voltage fell from 100 V to 16.7 V, which is a factor of K, while the charge stayed the same. That is exactly why the capacitance rose by the factor K. Outside the plates, the ideal field is zero throughout.
Example 5: A slab that only partly fills the gap
In the same capacitor (A = 0.02 m², d = 1 mm), a dielectric slab with K = 5 and thickness t = 0.5 mm is inserted, leaving a 0.5 mm air gap. Find the capacitance.
Check with fields: starting from Q = 17.7 nC, E_free = 1.0 × 10⁵ V/m in the air gap and E_free/K = 2.0 × 10⁴ V/m in the slab. So V = (10⁵)(0.5 mm) + (2 × 10⁴)(0.5 mm) = 50 + 10 = 60 V, and C = Q/V = 17.7 nC / 60 V = 295 pF. Both routes agree.
Check with fields: starting from Q = 17.7 nC, E_free = 1.0 × 10⁵ V/m in the air gap and E_free/K = 2.0 × 10⁴ V/m in the slab. So V = (10⁵)(0.5 mm) + (2 × 10⁴)(0.5 mm) = 50 + 10 = 60 V, and C = Q/V = 17.7 nC / 60 V = 295 pF. Both routes agree.
Example 6: Energy stored
(a) Using the three forms. The 100 µF capacitor of Example 1 holds Q = 1.2 mC at V = 12 V:
Check with fields: starting from Q = 17.7 nC, E_free = 1.0 × 10⁵ V/m in the air gap and E_free/K = 2.0 × 10⁴ V/m in the slab. So V = (10⁵)(0.5 mm) + (2 × 10⁴)(0.5 mm) = 50 + 10 = 60 V, and C = Q/V = 17.7 nC / 60 V = 295 pF. Both routes agree.
Example 6: Energy stored
(a) Using the three forms. The 100 µF capacitor of Example 1 holds Q = 1.2 mC at V = 12 V: U=1/2 CV^2 =7.2
(b) What a dielectric does to the energy. The 177 pF capacitor of Example 2, charged to 100 V, stores:
Now insert the mica (K = 6) in two different ways:
| Situation | What stays fixed | Energy afterwards |
|---|---|---|
| Battery stays connected | V = 100 V | U = K U₀ = 5.31 × 10⁻⁶ J (six times larger; the battery supplies the extra) |
| Battery disconnected first | Q = 17.7 nC | U = U₀ / K = 1.475 × 10⁻⁷ J (six times smaller; the field pulls the slab in) |
You can confirm the second row with Example 4: U = ½QV = ½(17.7 nC)(16.7 V) = 1.48 × 10⁻⁷ J.
Practice problems
- A 2.2 µF capacitor is charged to 50 V. Find the charge and the stored energy. (Answers: 110 µC, 2.75 mJ)
- Plates of area 0.05 m² are 2 mm apart in air. Find C₀. (Answer: about 221 pF)
- A dielectric with K = 4 fills that gap. Find the new capacitance, and the net field if E_free is 2 × 10⁵ V/m. (Answers: about 885 pF, 5 × 10⁴ V/m)
Frequently Asked Questions
What is the SI unit of capacitance?
The SI unit of capacitance is the farad (F), defined as one coulomb of charge stored per volt of potential difference (1 F = 1 C/V). A farad is very large, so practical capacitors are rated in microfarads (µF), nanofarads (nF) or picofarads (pF).
How does a capacitor work?
A capacitor has two conducting plates separated by an insulator. When it is connected to a battery, electrons move from one plate to the other, leaving equal and opposite charges +Q and −Q on the plates. This builds an electric field in the gap and a potential difference V across the plates, with Q = CV. Charging stops when V equals the battery voltage.
What does the capacitance of a parallel plate capacitor depend on?
Capacitance depends only on the geometry and the material between the plates: C = ε₀A/d for vacuum or air, and C = Kε₀A/d when a dielectric fills the gap. It rises with plate area A and with the dielectric constant K, and falls as the plate separation d grows. It does not depend on the charge or the voltage.
Why does a dielectric increase the capacitance?
The field of the plates polarises the dielectric, producing bound surface charges (−σ_ind and +σ_ind) on its faces. These create an induced field E_ind that opposes the free-charge field, so the net field inside falls to E_free/K. A weaker field means a smaller potential difference for the same stored charge, and since C = Q/V, the capacitance rises by the factor K.
What is the dielectric constant?
The dielectric constant K is the ratio of the capacitance with the dielectric to the capacitance with vacuum between the plates: K = C/C₀. It is also the ratio of the field without the dielectric to the net field with it, K = E₀/E = E_free/E_net. It has no units, equals 1 for vacuum, and is greater than 1 for every material.
What is the electric field outside a parallel plate capacitor?
For an ideal parallel plate capacitor, the field outside the plates is zero, because the fields of the two oppositely charged plates cancel there. Between the plates the field is uniform, E = σ/ε₀ without a dielectric, and E_free/K inside a dielectric. Near the edges the real field bulges outward (fringing), which the ideal model ignores.
How is energy stored in a capacitor?
Charging a capacitor requires work to move charge against the growing potential difference between the plates. That work is stored as electrical potential energy in the electric field between the plates, and it equals the area under the Q–V graph: U = ½QV = ½CV² = Q²/2C.
What happens to the stored energy when a dielectric is inserted?
It depends on what is held fixed. If the battery stays connected (V constant), the energy increases by a factor of K, because the battery supplies the extra. If the battery is disconnected first (Q constant), the energy decreases by a factor of K, and the difference is the work done by the field in pulling the dielectric into the gap.
Can a capacitor be charged beyond its voltage rating?
No. Every real dielectric has a maximum field it can withstand, called its dielectric strength. If the voltage pushes the field past that limit, the insulator breaks down, a spark or current passes through it, and the capacitor can be damaged or fail dangerously. This is why capacitors carry a voltage rating, such as 450 V on the 47 µF capacitor shown earlier.
A capacitor has two conducting plates separated by an insulator. When connected to a battery, charge moves from one plate to the other, leaving equal and opposite charges +Q and −Q. This creates an electric field between the plates and a potential difference V, with Q = CV. Charging stops when V equals the battery voltage.
The plates’ field polarises the dielectric, creating bound charges on its faces. These produce an induced field that opposes the original field, so the net field inside falls to E_free/K. A weaker field means a smaller voltage for the same charge, and since C = Q/V, the capacitance increases by the factor K.
Charging a capacitor requires work to move charge against the rising potential difference. This work is stored as electrical potential energy in the field between the plates and equals the area under the Q–V graph: U = ½QV = ½CV² = Q²/2C.
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