Capacitor

Capacitance: the ratio of the charge stored on a capacitor to the potential difference across it, $C=\dfrac{Q}{V}$. Its SI unit is the farad (F), $1\ \text{F}=1\ \text{C V}^{-1}$.

Energy stored in a capacitor: Energy is needed to move charge onto the plates against the potential difference already built up; this work is stored as electric potential energy.

Suppose the capacitor has capacitance $C$ and at some instant carries charge $q$, so the potential difference is $v=\dfrac{q}{C}$. The work needed to add a small charge $dq$ is

$dW=v\,dq=\frac{q}{C}\,dq$

Total work to charge it from $0$ to $Q$:

$W=\int_0^Q\frac{q}{C}\,dq=\frac{Q^2}{2C}$

This work is stored as energy $U$. Using $Q=CV$,

$U=\frac{Q^2}{2C}=\tfrac12CV^2=\tfrac12QV$

The energy is stored in the electric field between the plates, with energy density $\tfrac12\varepsilon_0E^2$.

The capacitor filled with mica has the larger capacitance.

With air (or vacuum) between the plates, $C_0=\dfrac{\varepsilon_0A}{d}$. With a dielectric of relative permittivity (dielectric constant) $K$, the capacitance is

$C=\frac{K\varepsilon_0A}{d}=KC_0$

The plates are identical in area $A$ and separation $d$, and for mica $K\approx5$ to $7$ while for air $K\approx1$. Hence $C_{\text{mica}}=K C_{\text{air}}>C_{\text{air}}$. The dielectric's polarisation reduces the net field and potential difference for the same charge, so more charge can be stored per volt.

In series, the charge on each capacitor is the same, and the potential differences add.

Equivalent capacitance:

$C=\frac{C_1C_2}{C_1+C_2}=\frac{5\times15}{5+15}=3.75\ \mu\text{F}$

Charge on each capacitor:

$Q=CV=3.75\times10^{-6}\times220=8.25\times10^{-4}\ \text{C}=825\ \mu\text{C}$

Potential across each:

$V_1=\frac{Q}{C_1}=\frac{825}{5}=165\ \text{V},\qquad V_2=\frac{Q}{C_2}=\frac{825}{15}=55\ \text{V}$

Check: $165+55=220$ V.

The figure with the capacitor values was not available, so the answer is given as the method; substitute the values from the diagram.

Equivalent capacitance:

• Capacitors in parallel: $C_p=C_1+C_2+\dots$
• Capacitors in series: $\dfrac1{C_s}=\dfrac1{C_1}+\dfrac1{C_2}+\dots$

First combine each parallel group, then combine the resulting capacitors in series (or the reverse, according to the circuit) until a single $C_{eq}$ remains.

Charge on each capacitor:

1. Total charge drawn from the source: $Q=C_{eq}V$.
2. Capacitors in series carry the same charge, equal to the charge on their equivalent: $Q_1=Q_2=Q$.
3. For a parallel group with voltage $V_p$ across it (found from $V_p=Q/C_p$), each capacitor's charge is $q=C\,V_p$, and these add to the charge of the group.

(a) The slope of a $Q$–$V$ graph is $\dfrac{Q}{V}$, which is the capacitance $C$ of the capacitor.

(b) The area under the graph (shaded) represents the energy stored in the capacitor, i.e. the work done in charging it ($\tfrac12 QV$).

(c) $E=\tfrac12 CV^2=\tfrac12\times(100\times10^{-6})\times(250)^2=\tfrac12\times10^{-4}\times62500=3.125\ \text{J}\approx3.1\ \text{J}$

(c) in the field between the plates

The energy $\tfrac12CV^2$ is stored in the electric field set up between the plates, with energy density $\tfrac12\varepsilon_0E^2$ per unit volume. It is not located on the plates themselves.

The effect depends on whether the battery stays connected.

• Battery disconnected (charge $Q$ constant): inserting a dielectric of constant $K$ raises the capacitance to $C'=KC$. The stored energy $U=\dfrac{Q^2}{2C}$ therefore falls to $U'=\dfrac{U}{K}$, decreasing. The missing energy is the work done by the field in pulling the slab in.
• Battery connected (potential $V$ constant): $C$ rises to $KC$, so $U=\tfrac12CV^2$ rises to $KU$. The extra energy is supplied by the battery.

Treat the cloud and earth as a parallel plate capacitor with $A=50\ \text{km}^2=5\times10^{7}\ \text{m}^2$, $d=1\ \text{km}=10^{3}\ \text{m}$ and $V=100\ \text{kV}=10^{5}\ \text{V}$.

Capacitance:

$C=\frac{\varepsilon_0A}{d}=\frac{8.85\times10^{-12}\times5\times10^{7}}{10^{3}}=4.43\times10^{-7}\ \text{F}$

Energy stored:

$U=\tfrac12CV^2=\tfrac12\times4.43\times10^{-7}\times(10^{5})^2\approx2.2\times10^{3}\ \text{J}$

The energy stored is about $2.2\ \text{kJ}$.

Two uses: (i) storing electric charge and energy (for example in a camera flash), and (ii) smoothing (filtering) the rectified output in power supplies, or tuning radio circuits together with an inductor.

(a) The capacitance of a parallel plate capacitor is the charge stored per unit potential difference between its plates:

$C=\frac{Q}{V}$

Its SI unit is the farad (F). For plates of area $A$ separated by $d$ in air, $C=\dfrac{\varepsilon_0A}{d}$.

Application: a capacitor is used to smooth the output of a rectifier (or to store charge in a camera flash, or to block d.c. while passing a.c.).

(b)(i) Two capacitors are joined in parallel, and this pair is in series with the third, so the combination is a mixed (series–parallel) combination.

Parallel pair:

$C_p=1000+1000=2000\ \mu\text{F}$

In series with the third capacitor:

$\frac{1}{C}=\frac{1}{2000}+\frac{1}{1000}=\frac{3}{2000}\ \Rightarrow\ C=\frac{2000}{3}\approx667\ \mu\text{F}$

Capacitance of a parallel plate capacitor is the ability of the capacitor to store electric charge per unit potential difference between its plates. If a charge $Q$ given to the plates produces a potential difference $V$ between them, then

$C = \frac{Q}{V}$

Its SI unit is the farad (F). For a parallel plate capacitor with plate area $A$ and separation $d$, $C = \dfrac{\varepsilon_0 A}{d}$.

One application: It is used in electric circuits for storing charge and energy, e.g. in a camera flash, in tuning circuits of radio receivers, and as a filter (to block d.c. and pass a.c.) in rectifier circuits.

(i) In Fig. 7.1 the capacitors $C_1$ and $C_2$ are joined side by side (their plates connected to the same two points), and this pair is then joined end to end with $C_3$. So the arrangement is a mixed (series–parallel) combination — a parallel combination in series with a single capacitor.

Calculation of effective capacitance

Capacitors $C_1$ and $C_2$ are in parallel, so their equivalent capacitance is

$C_p = C_1 + C_2 = 1000 + 1000 = 2000\ \mu\text{F}$

Now $C_p$ and $C_3$ are in series, so

$\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{C_3} = \frac{1}{2000} + \frac{1}{1000} = \frac{1+2}{2000} = \frac{3}{2000}\ \mu\text{F}^{-1}$

$C_{eq} = \frac{2000}{3} = 666.67\ \mu\text{F} \approx 667\ \mu\text{F}$

So the effective capacitance of the combination is about $6.67 \times 10^{-4}$ F.

The capacitance of a parallel plate capacitor,

$C=\frac{\varepsilon_0\varepsilon_r A}{d}$

depends on:

1. the area $A$ of the plates (directly proportional),
2. the separation $d$ between the plates (inversely proportional),
3. the dielectric constant $\varepsilon_r$ of the medium between the plates (directly proportional).

Charge on the first capacitor:

$Q=C_1V_1=6\times10^{-6}\times150=9\times10^{-4}\ \text{C}$

Connecting an uncharged capacitor $C_2$ in parallel shares this charge without loss, so $Q=(C_1+C_2)V$ with $V=90\ \text{V}$:

$C_1+C_2=\frac{Q}{V}=\frac{9\times10^{-4}}{90}=10\ \mu\text{F}$

$C_2=10-6=4\ \mu\text{F}$

The second capacitor has capacitance $4\ \mu\text{F}$.