Capacitance, Dielectrics & Energy

PHYPotential & Capacitance

A capacitor stores charge and electrostatic energy; its capacitance $C=Q/V$ depends only on geometry and the dielectric, not on the charge applied. Inserting a dielectric of constant $K$ multiplies the capacitance by $K$ .

ISC problems combine capacitors in series and parallel, compute stored charge and energy, and analyse common-potential energy loss when capacitors are connected together.

Definition of capacitance

C = \dfrac{Q}{V}

C

in farads (F),

Q

charge on a plate (C),

V

potential difference (V);

1\,\mu\text{F}=10^{-6}\,\text{F}

Parallel-plate capacitor

C = \dfrac{K\varepsilon_0 A}{d}

A

plate area (

\text{m}^2

d

separation (m),

K=1

for air/vacuum,

\varepsilon_0=8.85\times10^{-12}\,\text{F m}^{-1}

Derivation

Each plate carries surface density $\sigma = Q/A$ . The two plates' fields add, giving a uniform field in the gap (filled with dielectric $K$ ):
$E = \dfrac{\sigma}{K\varepsilon_0} = \dfrac{Q}{K\varepsilon_0 A}$
The potential difference across a gap of width $d$ is $V = Ed$ :
$V = E d = \dfrac{Q d}{K\varepsilon_0 A}$
Capacitance is $C = Q/V$ ; the charge $Q$ cancels:
$C = \dfrac{Q}{V} = \dfrac{K\varepsilon_0 A}{d}$

Series and parallel combination

\dfrac{1}{C_s}=\sum_i\dfrac{1}{C_i}, \qquad C_p=\sum_i C_i

series stores the same charge with smaller net

C

; parallel has the same voltage with larger net

C

Derivation

Series: every capacitor holds the same charge $Q$ , while the applied voltage splits, $V = V_1 + V_2 + \cdots$ . Using $V_i = Q/C_i$ and cancelling $Q$ :
$\dfrac{Q}{C_s} = \dfrac{Q}{C_1} + \dfrac{Q}{C_2} + \cdots \implies \dfrac{1}{C_s} = \sum_i \dfrac{1}{C_i}$
Parallel: every capacitor has the same voltage $V$ , while the charges add, $Q = Q_1 + Q_2 + \cdots$ . Using $Q_i = C_i V$ and cancelling $V$ :
$C_p V = C_1 V + C_2 V + \cdots \implies C_p = \sum_i C_i$

Energy stored

U = \dfrac{1}{2}CV^2 = \dfrac{1}{2}QV = \dfrac{Q^2}{2C}

use the form matching the known quantities;

U

in joules

Derivation

Charging happens in steps: adding $dq$ when the capacitor already sits at potential $V' = q/C$ needs work:
$dW = V'\,dq = \dfrac{q}{C}\,dq$
Integrate from $0$ to the final charge $Q$ :
$U = \int_0^{Q} \dfrac{q}{C}\,dq = \dfrac{Q^2}{2C}$
Using $Q = CV$ gives the three equivalent forms:
$U = \dfrac{Q^2}{2C} = \dfrac{1}{2}QV = \dfrac{1}{2}CV^2$

Energy density of the field

u = \dfrac{1}{2}\varepsilon_0 E^2

u

is energy per unit volume (

\text{J m}^{-3}

) stored in vacuum; replace

\varepsilon_0

K\varepsilon_0

in a dielectric

Derivation

Take a vacuum parallel-plate capacitor: $C = \varepsilon_0 A/d$ and $V = Ed$ . Its stored energy is:
$U = \dfrac{1}{2}CV^2 = \dfrac{1}{2}\left(\dfrac{\varepsilon_0 A}{d}\right)(Ed)^2 = \dfrac{1}{2}\varepsilon_0 E^2\,(Ad)$
The field fills the volume $Ad$ between the plates, so the energy per unit volume is:
$u = \dfrac{U}{Ad} = \dfrac{1}{2}\varepsilon_0 E^2$
The result is general — energy resides in the field itself, wherever a field exists.

Energy lost on sharing charge

\Delta U = \dfrac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)}

common potential

V=\dfrac{C_1 V_1+C_2 V_2}{C_1+C_2}

; energy is always lost (as heat/radiation) if

V_1\neq V_2

Derivation

Charge is conserved when the capacitors are connected, so they settle at a common potential:
$V = \dfrac{Q_1 + Q_2}{C_1 + C_2} = \dfrac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$
The loss is $\Delta U = U_i - U_f$ with $U_i = \tfrac{1}{2}C_1 V_1^2 + \tfrac{1}{2}C_2 V_2^2$ and $U_f = \tfrac{1}{2}(C_1+C_2)V^2$ . Substituting $V$ and simplifying:
$\Delta U = \dfrac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)}$
The numerator is a square, so $\Delta U \ge 0$ : energy is always lost (as heat in the wires) unless $V_1 = V_2$ .

Capacitance depends only on geometry and dielectric — charging it more raises $Q$ and $V$ together but leaves $C$ unchanged.
Series capacitors all carry the same charge; parallel capacitors all share the same potential difference.
Inserting a dielectric increases $C$ by factor $K$ , which raises the stored charge if the battery stays connected ( $V$ fixed) or lowers the voltage if the battery is disconnected ( $Q$ fixed).
With the battery disconnected, $Q$ is fixed; increasing the plate separation $d$ lowers $C$ , so $U=Q^2/2C$ increases (work done against attraction).
Energy sharing always loses energy when the two initial potentials differ — charge is conserved but energy is not.
The common potential equals total charge divided by total capacitance, $V=Q_{total}/(C_1+C_2)$ .
Field energy can be viewed as stored in the space between the plates with density $u=\tfrac{1}{2}\varepsilon_0 E^2$ .

Where the marks go

Treating the series formula like resistance addition — for series you add reciprocals, $1/C_s=\sum 1/C_i$ , giving a value smaller than the smallest capacitor.
Using $U=\tfrac{1}{2}QV$ with the wrong pairing of variables; pick the energy form that uses quantities you actually know.
Assuming energy is conserved when two capacitors are connected — it is not; only charge is conserved.
Forgetting $A$ must be in $\text{m}^2$ ( $100\,\text{cm}^2=10^{-2}\,\text{m}^2$ ) and $d$ in metres in $C=K\varepsilon_0 A/d$ .

How the board asks it

Numericalseries and parallel combination, stored charge and energy
Three capacitors of $2\,\mu F$ , $3\,\mu F$ and $6\,\mu F$ are connected in series across a $100\,V$ supply. Calculate the equivalent capacitance, the charge on each capacitor and the total energy stored.
Numerical $C=K\varepsilon_0 A/d$ with a dielectric slab
A parallel-plate capacitor has plates of area $100\,cm^2$ separated by $2\,mm$ . Calculate its capacitance when the gap is completely filled with a dielectric of constant $K=5$ . $(\varepsilon_0=8.85\times10^{-12}\,F\,m^{-1})$
Numericalcommon-potential energy loss on sharing charge
A $4\,\mu F$ capacitor charged to $200\,V$ is connected across an uncharged $2\,\mu F$ capacitor. Calculate the common potential and the energy lost in the process.
Derive / proveenergy stored and energy density of the field
Obtain an expression for the energy stored in a charged parallel-plate capacitor, and hence show that the energy density of the electric field between the plates is $u=\tfrac{1}{2}\varepsilon_0 E^2$ .
Distinguishbattery connected ( $V$ fixed) vs disconnected ( $Q$ fixed)
When a dielectric slab is inserted between the plates of a capacitor, distinguish between the changes in charge, potential difference and stored energy in the two cases where the battery remains connected and where it has been disconnected.
Give reasons $U=Q^2/2C$ with $Q$ fixed
A parallel-plate capacitor is charged and then disconnected from the battery. Explain why the stored energy increases when the separation between its plates is increased.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.