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Capacitors and energy stored

A capacitor is two plates and a gap, rated by how much charge each volt of pd parks on them. Fill the gap with the right material and the rating climbs; charge it up and the energy it holds is a triangle on a graph, growing as the square of the voltage.

Year 13AQA 3.7.4.1, 3.7.4.2, 3.7.4.3

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.
  • Find stored energy from the Q–V graph and use all three energy forms.

WHAT YOU PROBABLY THINK

A capacitor is a small battery.

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 matches the supply. The capacitance rates how much charge arrives per volt:

C = QVON YOUR DATA SHEET
A charged capacitor: equal and opposite charge on two plates of area A, a distance d apart, with a uniform field between+Q−Qdplate area AC = Q/V: how much charge each volt of pd stores
FIG. 1The charged capacitor: +Q and −Q on plates of area A a distance d apart, with a uniform field in the gap.

measured in farads, coulombs per volt. One farad is an enormous rating, so real components live in microfarads, nanofarads and picofarads, and the prefix arithmetic from Year 12 earns its keep here. Note the bookkeeping: Q names the charge on one plate, and the capacitor as a whole stays neutral, +Q and −Q cancelling exactly.

Geometry, and the dielectric

For the parallel-plate design the rating follows from the geometry and the filling:

C = Aε0εrdON YOUR DATA SHEET

bigger plates and a smaller gap store more charge per volt, and the factor εr, the relative permittivity or dielectric constant of the insulating filling, multiplies the whole rating.

A polar dielectric between the plates: with no field the molecules point every way; switch the field on and they rotate to line upno field: randomfield on: alignedcyan end positive, grey end negativealignment weakens the field, so C rises
FIG. 2A polar dielectric: molecules point every way with no field, then rotate into alignment when the field switches on.

The spec asks you to describe why, at the level of one 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 point every way; switch the field on and each one feels a turning effect and rotates to align with the field, negative ends toward the positive plate. The 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.

The energy, and why the half

Charge against pd is a straight line through the origin, and the energy stored is the triangular area beneath itVQVQarea = ½QV = energy storedgradient = C; every extra coulomb costs more
FIG. 3Charge against pd is a straight line, and the stored energy is the triangle beneath it: half Q V.

Because Q = CV, the graph of charge against pd is a straight line through the origin with gradient C, and the energy stored is the area under it. The first coulomb arrives with the plates nearly uncharged and costs almost nothing; the last is pushed on against the full pd. Averaging over the filling gives the triangle's area:

E = 12QV = 12CV2 = 12Q2CON YOUR DATA SHEET
Energy grows as the square of the voltage: doubling V quadruples the energy a capacitor storesat Vat 2VE4EE = ½CV²: the square does the work
FIG. 4Half C V squared at work: doubling the voltage quadruples the stored energy.

with the other two forms reached by substituting Q = CV. The square in ½CV2 means doubling the charging voltage quadruples the energy. The lie above falls to a comparison: a battery makes energy chemically and holds its pd nearly steady while it lasts, whereas a capacitor merely stores energy in its field, its pd collapsing as the charge drains, and it can dump its whole store in milliseconds, which is exactly why camera flashes use one.

THE EXAM BIT

  • C = Q/V defines the rating; quote Q as the magnitude of the charge on one plate. The pair together is neutral.
  • The polar-molecule description earns its marks in sequence: two charged ends, a turning effect in the field, rotation into alignment, and the resulting rise in C. Tell it as that story.
  • In the plate formula every symbol moves C the way intuition says: A up, d down, εr up. Check limits before trusting algebra.
  • Energy questions reward choosing the form that matches the given data: ½QV when both are known, ½CV2 from the rating and the voltage, ½Q2/C from the charge alone.
  • The area-under-the-graph justification for the half is itself examinable: each extra coulomb is pushed on against a growing pd, so the average pd during charging is V/2.

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 and energy it stores at 12 V.

Show a hint

Geometry first, then Q = CV, then the energy form that uses what you now hold.

Show the answer

C = Aε0εr/d = (0.020 × 8.85 × 10−12 × 2.5) / 1.0 × 10−3 = 4.4 × 10−10 F, about 0.44 nF.

Q = CV = 4.4 × 10−10 × 12 = 5.3 × 10−9 C.

E = 12CV2 = 0.5 × 4.4 × 10−10 × 144 = 3.2 × 10−8 J: tiny numbers, which is why the farad so rarely appears without a prefix.

Capacitance is charge parked per volt; the dielectric raises it.

Energy is the triangle: ½QV, growing as the voltage squared.

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