Physics › Capacitance › Capacitors and the dielectric
Capacitors and the dielectric
Capacitance is the charge stored on one plate per unit potential difference, C = Q/V, and it is measured in farads, one coulomb per volt. For parallel plates C = Aε₀ε_r/d. A dielectric raises the capacitance, because its polar molecules rotate into line and weaken the field in the gap.
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Capacitors and energy stored, part 1 of 2. Part 2 is Energy stored and capacitor combinations.
Builds on Coulomb's law and electric field strength and Electric potential.
IN THIS TOPIC
- Define capacitance with C = Q/V and use the parallel-plate formula.
- Describe how a polar dielectric molecule rotates in the field, and why that raises C.
COMMON MISCONCEPTION
An insulating filling between the plates weakens a capacitor.
The opposite: the dielectric's polarised molecules oppose the field between the plates, lowering the pd the same charge produces, so C = Q/V rises by the relative permittivity.
What capacitance measures
Connect two conducting plates across a supply and charge flows until one plate holds +Q, the other −Q, and the pd across the pair equals the supply pd. Capacitance is the charge stored per unit potential difference.
measured in farads, coulombs per volt. One farad is a very large capacitance, so practical components are rated in microfarads, nanofarads and picofarads; convert the prefix before calculating. By convention Q is the magnitude of the charge on one plate, and the capacitor as a whole is neutral because +Q and −Q cancel exactly.
CIE widens that definition past the two-plate component. Any isolated conductor has a capacitance of its own, defined the same way as the charge on it per unit potential, except that the potential is now measured against zero at infinity rather than across a gap to a facing plate. For a sphere the electric potential lesson has already done the work. A conducting sphere of radius r carrying charge Q sits at , so dividing the charge by the potential leaves
so the capacitance depends only on the radius. A sphere of the Earth's radius has a capacitance of about 700 μF, which indicates how large one farad is. The other boards define capacitance for the two-plate component only.
WORKED EXAMPLE
How much charge a flash stores
A camera-flash capacitor of 2200 μF is charged to 5.0 V. Find the charge stored.
Q = CV = 2200 × 10−6 × 5.0 = 0.011 C.
That charge corresponds to about 7 × 1016 electrons transferred to one plate.
The calculation is the definition rearranged: capacitance is the charge stored per unit pd, so the charge stored is the capacitance multiplied by the pd.
Geometry, and the dielectric
For a parallel-plate capacitor the capacitance depends on the plate area, the plate separation and the material between the plates.
Larger plates and a smaller separation give a greater charge stored per unit pd, and the factor εr, the relative permittivity or dielectric constant of the insulating material between the plates, multiplies the capacitance.
Describe the drawing
Between the plates, polar molecules point every which way while the field is off. Switch it on and they swing into line, positive ends toward the negative plate, with a little overshoot as they settle; switch it off and thermal jostling scatters them again. That alignment opposes the applied field inside the material, which is why a dielectric raises a capacitor's capacitance.
The explanation required is at the level of a single molecule. A polar molecule carries a slight positive charge at one end and a slight negative charge at the other. With no field the molecules are randomly oriented. When the field is applied, each molecule experiences a turning effect and rotates to align with the field, negative ends towards the positive plate. That layer of aligned charge partly cancels the field in the gap, so a smaller pd appears for the same stored charge, and C = Q/V rises.
Rotation of permanent dipoles is the mechanism to describe, but it is not the only one. A non-polar dielectric such as polythene has no permanent dipoles, so the applied field displaces each molecule's electron cloud slightly and induces a dipole. The outcome is the same in both cases: the aligned dipoles leave a layer of bound charge on each face, that layer produces a field opposing the applied field, and the capacitance increases.
The energy, and why the half has moved to the next part of this lesson.
Capacitors in series and in parallel, for CIE and OCR has moved to the next part of this lesson.
ASSESSMENT FOCUS
- C = Q/V defines the rating, and Q means the magnitude of the charge on one plate. The pair together is neutral.
- CIE alone applies the definition to an isolated conductor as well, where the potential is measured from zero at infinity, and an isolated sphere of radius r has capacitance 4πε0r. Give the definition in that form and the sphere formula follows from V = Q/(4πε0r) in one line.
- A farad is enormous, so real ratings arrive in μF, nF and pF. Convert the prefix before the arithmetic, and expect an answer that carries one too.
- Give the polar-molecule description in sequence: the molecule has two oppositely charged ends, the field exerts a turning effect on it, it rotates into alignment, and the capacitance rises as a result. Each step carries credit.
- In the plate formula, C increases with A, decreases with d and increases with εr. Check an answer against those three dependences.
CHECK YOURSELF
A capacitor has plates of area 0.020 m² separated by 1.0 mm, with a dielectric of relative permittivity 2.5. Find its capacitance and the charge it stores at 12 V, and state and explain what removing the dielectric does to the capacitance.
Show a hint
Use the plate formula first, then Q = CV, then apply the polar-molecule mechanism in reverse.
Show the answer
= (0.020 × 8.85 × 10−12 × 2.5) / (1.0 × 10−3) = 4.4 × 10−10 F, about 0.44 nF.
= 4.4 × 10−10 × 12 = 5.3 × 10−9 C. Values of this size are why capacitances are normally quoted with a prefix.
Removing the dielectric divides C by εr = 2.5, to 1.8 × 10−10 F. No aligned dipoles are left to oppose the field in the gap, so the full pd returns for the same charge, and C = Q/V falls.
Capacitance is the charge stored per unit potential difference: C = Q/V.
A dielectric between the plates increases it.
In the plate formula A sits on top, d underneath, and ε_r multiplies the rating.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the capacitors and the dielectric questions page.
WHERE TO GO NEXT
- Gradients and areas under graphs is the maths this lesson leans on, worked through step by step.
USED LATER IN
CHECK YOUR PROGRESS
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- Define capacitance with C = Q/V and use the parallel-plate formula.
- Describe how a polar dielectric molecule rotates in the field, and why that raises C.
Open the full revision checklist to track your progress across the whole unit.