The Parallel Plate Capacitor

The derivation CBSE sets

Two large plane parallel conducting plates, each of area $A$, separated by $d$, with vacuum (or air) between them. Plate 1 carries $+Q$ and plate 2 carries $-Q$, so the surface densities are $\sigma=\dfrac{Q}{A}$ and $-\sigma$. Because $d^{2}\ll A$, each plate may be treated as an infinite sheet with field $\dfrac{\sigma}{2\varepsilon_0}$ on either side.

Region I, above plate 1:

$$E_{\text{I}}=\dfrac{\sigma}{2\varepsilon_0}-\dfrac{\sigma}{2\varepsilon_0}=0$$

Region III, below plate 2:

$$E_{\text{III}}=\dfrac{\sigma}{2\varepsilon_0}-\dfrac{\sigma}{2\varepsilon_0}=0$$

Region II, between the plates, where the two fields point the same way and add:

$$E_{\text{II}}=\dfrac{\sigma}{2\varepsilon_0}+\dfrac{\sigma}{2\varepsilon_0}=\dfrac{\sigma}{\varepsilon_0}=\dfrac{Q}{\varepsilon_0 A}$$

directed normal to the plates, from the positive plate to the negative plate. So the field is localised in the gap and uniform there.

The field is uniform, so the potential difference is field times distance:

$$V=E d=\dfrac{Qd}{\varepsilon_0 A}$$

$$C=\dfrac{Q}{V}=\dfrac{\varepsilon_0 A}{d}$$

which, as expected, depends only on the geometry.

Fringing

For plates of finite area the field lines bulge outwards near the edges — the fringing of the field — and $\sigma$ is not exactly uniform right at the rim. For $d^{2}\ll A$ these effects are negligible in the region well away from the edges, and the formula above holds there.

How big is a farad

• With $A=1\ \text{m}^{2}$ and $d=1\ \text{mm}$: $C=\dfrac{8.85\times10^{-12}\times1}{10^{-3}}=8.85\times10^{-9}\ \text{F}$, only about $9\ \text{nF}$.
• To get $C=1\ \text{F}$ with $d=1\ \text{cm}$ needs $A=\dfrac{Cd}{\varepsilon_0}\approx10^{9}\ \text{m}^{2}$, a square plate roughly $30\ \text{km}$ on a side.
• Worked pattern: $A=6\times10^{-3}\ \text{m}^{2}$, $d=3\ \text{mm}$ gives $C=\dfrac{8.85\times10^{-12}\times6\times10^{-3}}{3\times10^{-3}}=17.7\times10^{-12}\ \text{F}$, and on a $100\ \text{V}$ supply $Q=CV=17.7\times10^{-10}\ \text{C}$.

Charging across a dc battery — the explanation CBSE marks

• On connection, a momentary current flows: electrons accumulate on the plate joined to the negative terminal and an equal positive charge appears on the other plate. No charge crosses the insulating gap.
• As charge builds up, the potential difference across the plates grows and opposes the arrival of further charge, so the current falls progressively.
• Charging stops — the current becomes zero — when the potential difference across the plates equals the terminal voltage of the battery. The capacitor is then fully charged, with $Q=CV$.

Removing one plate

The field between two oppositely charged plates is $\dfrac{\sigma}{\varepsilon_0}$, twice the field $\dfrac{\sigma}{2\varepsilon_0}$ of a single plate. So if one plate is removed, the field, and hence the force $F=qE$ on a charge between the plates, falls to half: the answer is $F/2$.

Diagram to describe in words

Two horizontal plates, upper marked $+\sigma$ and lower marked $-\sigma$, separated by $d$, with the plate area $A$ labelled. Draw closely spaced downward arrows filling the gap for the uniform field $\dfrac{\sigma}{\varepsilon_0}$, and mark regions I above and III below with $E=0$. At the two ends, show a couple of arrows curving outward to indicate fringing.

Standard results