Chapter 2 · Electrostatic Potential and Capacitance

2.3 Conductors, Dielectrics and the Parallel-Plate Capacitor

Inside a conductor in equilibrium $E=0$ and the whole conductor is at one potential; a dielectric only weakens a field, by its dielectric constant $K$. Two conductors facing each other make a capacitor, and for parallel plates $C=\dfrac{K\varepsilon_0A}{d}$.

If lightning strikes a car, the people inside are almost always unhurt: the huge charge stays on the metal skin and the field inside the car stays close to zero. The same idea protects the signal in a television cable, whose inner wire is wrapped in a metal braid, and keeps the delicate circuits of an MRI scanner inside a metal-lined room. Metals behave this way because their free electrons rearrange, in a tiny fraction of a second, until they cancel any field inside the metal.

Insulators — dielectrics — cannot cancel a field, but they weaken it: their molecules become small dipoles that line up against the field. Both kinds of material meet in the capacitor, two conductors separated by a dielectric, which stores charge (and, as Section 2.4 shows, energy). This section sets out the rules for conductors in electrostatics, explains polarisation of a dielectric and the dielectric constant, defines capacitance, and derives the capacitance of the most important design, two parallel plates, with and without a dielectric between them.

Electrostatics of Conductors

A metal contains a huge number of free electrons — about $10^{29}$ per cubic metre in copper — that can move anywhere within it. Put a metal in an electric field and these electrons are pushed against the field. They pile up on one side of the metal and leave positive ions exposed on the other side, and these induced charges make a field of their own, opposing the applied one. The electrons keep moving until the two fields cancel exactly inside the metal, because as long as there is any field inside, the free electrons feel a force and move. In a good conductor this takes far less than a microsecond. Once it is over, the conductor is in electrostatic equilibrium, and the following results hold.

📐 Result 2.10 — A conductor in electrostatic equilibrium

For a conductor whose charges are at rest (whether it is charged or neutral, and whether or not an external field is present):

  1. The field inside the material is zero.
  2. Just outside, the field is normal to the surface. A tangential component would drive charges along the surface.
  3. There is no excess charge inside; any excess charge resides on the surface. (Apply Gauss's law to a tiny closed surface anywhere inside the metal: $E=0$ on it, so the charge it encloses is zero.)
  4. The potential is the same everywhere in the conductor and on its surface: with $E=0$ inside and no tangential $E$ on the surface, no work is done in moving a charge anywhere in or on it. The surface is an equipotential.
  5. The field just outside is $\mathbf E=\dfrac{\sigma}{\varepsilon_0}\,\hat{\mathbf n}$, where $\sigma$ is the local surface charge density and $\hat{\mathbf n}$ the outward normal. (A Gaussian pillbox half inside the metal has flux $E\,\delta A$ through its outer face only, and encloses $\sigma\,\delta A$.)
  6. Electrostatic shielding. In an empty cavity inside a conductor the field is zero, whatever charges or fields are outside — even if the conductor itself is charged.

Result 6 explains the car in a thunderstorm and the screened cable. It is not that the metal "blocks" the field: the charges on the outer surface of the conductor rearrange so that their field and the external field cancel everywhere inside the outer surface, including the cavity. The reverse is not true: a charge inside a cavity does affect the outside. A charge $+q$ placed in the cavity of a neutral conductor induces $-q$ on the cavity wall and leaves $+q$ on the outer surface, whose field reaches out into the world (unless the conductor is earthed).

Result 5 has a practical consequence. On a conductor of irregular shape, all of the surface is at one potential, but the charge density is not uniform — it is greatest where the surface curves most sharply. Two spheres of different radii joined by a long wire show why (Example 2). At a sharp point the surface field can be strong enough to ionise the air, which is how a lightning conductor leaks charge away gently, and why high-voltage equipment is built with smooth, rounded ends.

✏️ Example 1 — A charged metal sphere
A metal sphere of radius $10$ cm on an insulating stand carries a charge of $+20$ nC. Find (a) the surface charge density, (b) the field just outside the surface, (c) the potential of the sphere, and (d) the field and potential at its centre.
(a) Density The charge spreads uniformly over the outer surface (by symmetry): $\sigma=\dfrac{Q}{4\pi R^2}=\dfrac{2.0\times10^{-8}}{4\pi(0.10)^2}\approx1.59\times10^{-7}\ \text{C/m}^2$.
(b) Field $E=\dfrac{\sigma}{\varepsilon_0}=\dfrac{1.59\times10^{-7}}{8.85\times10^{-12}}\approx1.8\times10^{4}$ N/C, normal to the surface, outwards. Check: $\dfrac{kQ}{R^2}=\dfrac{(9.0\times10^9)(2.0\times10^{-8})}{0.010}=1.8\times10^4$ N/C ✓.
(c) Potential $V=\dfrac{kQ}{R}=\dfrac{180}{0.10}=1.8\times10^{3}$ V.
(d) Centre $E=0$ (inside a conductor); $V=1.8\times10^3$ V, the same as on the surface.
✏️ Example 2 — Why charge crowds onto sharp points
Two metal spheres of radii $2.0$ cm and $6.0$ cm, far apart, are joined by a long thin wire, and a total charge of $40$ nC is shared between them. Find (a) the charge on each sphere, (b) their common potential, and (c) the ratio of their surface charge densities and the field at each surface.
Concept Joined by a wire, the spheres form one conductor and must be at the same potential. Far apart, each one's potential is just $kq/R$ from its own charge.
(a) Charges $\dfrac{kq_1}{R_1}=\dfrac{kq_2}{R_2}\Rightarrow\dfrac{q_1}{q_2}=\dfrac{R_1}{R_2}=\dfrac13$. With $q_1+q_2=40$ nC: $q_1=10$ nC, $q_2=30$ nC. The larger sphere takes more charge.
(b) Potential $V=\dfrac{(9.0\times10^9)(1.0\times10^{-8})}{0.020}=4.5\times10^3$ V (and $\dfrac{(9.0\times10^9)(3.0\times10^{-8})}{0.060}$ gives the same ✓).
(c) Densities $\sigma=\dfrac{q}{4\pi R^2}\propto\dfrac{R}{R^2}=\dfrac1R$, so $\dfrac{\sigma_1}{\sigma_2}=\dfrac{R_2}{R_1}=3$. Surface fields $E=\dfrac{V}{R}$: $E_1=\dfrac{4500}{0.020}=2.25\times10^5$ N/C and $E_2=\dfrac{4500}{0.060}=7.5\times10^4$ N/C. The smaller sphere — the more sharply curved surface — has the larger charge density and the stronger field.
✏️ Example 3 — A charge inside a metal shell
A neutral metal spherical shell has inner radius $8.0$ cm and outer radius $10$ cm. A point charge $+5.0$ nC is held at its centre. Find (a) the charges on the inner and outer surfaces of the shell, (b) the field at $5.0$ cm, $9.0$ cm and $15$ cm from the centre, and (c) the potential of the shell.
(a) Induced charges A Gaussian sphere of radius $9.0$ cm lies inside the metal, where $E=0$, so it encloses zero charge: the inner surface carries $-5.0$ nC. The shell is neutral, so the outer surface carries $+5.0$ nC, spread uniformly.
(b) Fields At $5.0$ cm (in the cavity): only the point charge is enclosed, $E=\dfrac{45}{(0.050)^2}=1.8\times10^4$ N/C outwards. At $9.0$ cm (in the metal): $E=0$. At $15$ cm: enclosed charge $+5.0-5.0+5.0=+5.0$ nC, $E=\dfrac{45}{(0.15)^2}=2.0\times10^3$ N/C outwards.
(c) Potential Outside, everything acts as $+5.0$ nC at the centre, so at the outer surface $V=\dfrac{45}{0.10}=450$ V; the whole shell is at $450$ V.
Note If the shell is earthed, the $+5.0$ nC on its outer surface flows away; the field outside becomes zero while the field in the cavity is unchanged. Earthed shells are how sensitive instruments are screened from charges inside equipment.

Dielectrics and Polarisation

A dielectric — glass, mica, paper, plastic, pure water, air — has no free charges, so it cannot cancel a field. But its molecules respond in one of two ways.

Non-polar molecules ($\mathrm{O_2}$, $\mathrm{N_2}$, $\mathrm{H_2}$, $\mathrm{CO_2}$) have their centres of positive and negative charge at the same point, so their dipole moment is zero. An external field pulls the two centres slightly apart and gives each molecule a small induced dipole moment along the field. Polar molecules ($\mathrm{H_2O}$, $\mathrm{HCl}$, $\mathrm{NH_3}$) have a permanent dipole moment, but in a sample their directions are random because of thermal motion, so the total is zero. An external field exerts a torque on each of them (Section 1.2) and lines them up partly — only partly, because thermal jostling keeps disturbing them (Exercise III.6 of Section 2.2 shows how weak the alignment is).

Either way, the material acquires a net dipole moment along the field. Inside it, neighbouring $+$ and $-$ ends cancel; but on the two faces of a slab perpendicular to the field, a thin layer of uncancelled charge appears: negative on the face where the field enters, positive where it leaves. These bound (polarisation) charges cannot leave their molecules, but their field points against the external field inside the slab, and weakens it.

📘 Definition 2.4 — Electric polarisation, susceptibility and dielectric constant
The electric polarisation $\mathbf P$ of a dielectric is its dipole moment per unit volume (unit C/m²). For most materials, in fields that are not too strong, it is proportional to the field inside the material: $\mathbf P=\chi_e\mathbf E$, where $\chi_e$ is the electric susceptibility of the material. A uniformly polarised slab carries bound surface charge densities $\pm\sigma_p$ on its faces, with $\sigma_p=P$.
The dielectric constant $K$ of the material is the factor by which it reduces a field in which it is placed (filling the whole region between the charges that produce the field): $E=\dfrac{E_0}{K}$, with $$K=1+\frac{\chi_e}{\varepsilon_0}=\frac{\varepsilon}{\varepsilon_0}\ge1 .$$ $\varepsilon=K\varepsilon_0$ is the permittivity of the medium. $K$ is the same number as the relative permittivity $\varepsilon_r$ met in Coulomb's law (Section 1.1): vacuum $1$, air $1.0006$, paper about $3$, glass $5$–$10$, mica about $6$, water $80$. For a metal $K\to\infty$, since $E=0$ inside.

Where does $K=1+\chi_e/\varepsilon_0$ come from? Put a slab in a uniform field $E_0$ normal to its faces. The bound charges $\pm\sigma_p$ on the faces act like a pair of oppositely charged sheets and produce a field $\sigma_p/\varepsilon_0=P/\varepsilon_0$ inside the slab, opposing $E_0$. So $E=E_0-\dfrac{P}{\varepsilon_0}=E_0-\dfrac{\chi_eE}{\varepsilon_0}$, which gives $E_0=E\left(1+\dfrac{\chi_e}{\varepsilon_0}\right)=KE$. A conductor behaves like the limit $K\to\infty$: its induced charge density is $\varepsilon_0E_0$, just enough to cancel $E_0$ completely.

✏️ Example 4 — Bound charge on a dielectric slab
A large slab of a dielectric with $K=3.0$ is placed in a uniform field $E_0=3.0\times10^5$ V/m perpendicular to its faces. Find (a) the field inside the slab, (b) the polarisation and the bound surface charge density, and (c) the electric susceptibility. (d) What would the induced surface charge density be if the slab were made of metal?
(a) Field $E=\dfrac{E_0}{K}=\dfrac{3.0\times10^5}{3.0}=1.0\times10^5$ V/m, in the same direction as $E_0$.
(b) Polarisation $E=E_0-\dfrac{P}{\varepsilon_0}\Rightarrow P=\varepsilon_0(E_0-E)=(8.85\times10^{-12})(2.0\times10^5)\approx1.8\times10^{-6}\ \text{C/m}^2$, and $\sigma_p=P\approx1.8\times10^{-6}\ \text{C/m}^2$ — negative on the face where the field enters, positive on the other.
(c) Susceptibility $\chi_e=\varepsilon_0(K-1)=2\varepsilon_0\approx1.8\times10^{-11}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$ (check: $\chi_eE=(1.77\times10^{-11})(1.0\times10^5)=1.77\times10^{-6}=P$ ✓).
(d) Metal The field inside must vanish: $\sigma=\varepsilon_0E_0=(8.85\times10^{-12})(3.0\times10^5)\approx2.7\times10^{-6}\ \text{C/m}^2$ — larger than for the dielectric, which only partly cancels the field. Interactive 1 shows both cases.
Interactive 1 — A Conductor and a Dielectric in a Uniform Field

A slab is placed between two large charged plates that make a uniform field $E_0$ (to the right). Press Play to switch the field on and watch the slab respond. In a conductor free electrons drift left until the induced charges $\pm\sigma$ cancel the field inside completely. In a dielectric the molecules only stretch and turn into small dipoles, so the bound charges cancel just part of the field: $E=E_0/K$. The strip at the bottom plots $E$ across the gap.

3.03.0
field outside $E_0$—
field inside the slab—
induced / bound $\sigma$—
response complete—

Capacitors and Capacitance

Give an isolated conductor a charge $Q$ and its potential rises in proportion: $V=kQ/R$ for a sphere. Double the charge and every field and potential doubles. So the ratio $Q/V$ is a fixed number for the conductor, set by its size and shape. A much larger ratio can be had with two conductors close together, carrying equal and opposite charges: each one's charge pulls the other's potential towards its own, so a large charge can be stored at a small potential difference.

📘 Definition 2.5 — Capacitor and capacitance
A capacitor is a pair of conductors, insulated from each other, carrying charges $+Q$ and $-Q$. (Connecting them to the terminals of a battery charges them this way: the battery moves charge $Q$ from one conductor to the other.) Its capacitance is the ratio of the charge to the potential difference $V$ between the conductors: $$C=\frac{Q}{V}.$$ $C$ depends only on the sizes, shapes and separation of the conductors and on the medium between them — not on $Q$ or $V$. The SI unit is the farad: $1\ \text{F}=1\ \text{C/V}$. A farad is enormous; practical capacitors are measured in $\mu\text{F}$ ($10^{-6}$ F), nF ($10^{-9}$ F) and pF ($10^{-12}$ F). For a single isolated conductor the second "plate" is taken to be at infinity: an isolated sphere of radius $R$ has $C=\dfrac{Q}{kQ/R}=4\pi\varepsilon_0R$.

The parallel-plate capacitor. Take two flat metal plates, each of area $A$, a small distance $d$ apart, carrying charges $+Q$ and $-Q$. The charge spreads over the facing (inner) surfaces with densities $\pm\sigma=\pm Q/A$. Each plate is a large charged sheet with field $\sigma/2\varepsilon_0$ (Section 1.3). Between the plates the two fields point the same way (from $+$ to $-$) and add; outside they point in opposite directions and cancel: $$E_\text{between}=\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{2\varepsilon_0}=\frac{\sigma}{\varepsilon_0}=\frac{Q}{\varepsilon_0A},\qquad E_\text{outside}=0 .$$ The field between the plates is uniform, so the potential difference is $V=Ed=\dfrac{Qd}{\varepsilon_0A}$, and $C=Q/V$ gives the result. We have ignored the bending of the field near the edges (fringing), which is small when $d$ is much less than the size of the plates.

📐 Result 2.11 — Capacitance of a parallel-plate capacitor

Two parallel plates of area $A$ separated by a distance $d$ in vacuum (or air), with $d$ much smaller than the plate dimensions, have capacitance $$C_0=\frac{\varepsilon_0A}{d}.$$ The field between the plates is uniform, $E=\dfrac{\sigma}{\varepsilon_0}=\dfrac{V}{d}$, and the field outside is zero. Since $\varepsilon_0=Cd/A$, the unit of $\varepsilon_0$ can be written F/m: $\varepsilon_0=8.85\times10^{-12}$ F/m.

✏️ Example 5 — An air-filled parallel-plate capacitor
A capacitor has square plates of side $10$ cm, $2.0$ mm apart in air, and is connected to a $100$ V battery. Find (a) its capacitance, (b) the charge on each plate, (c) the field between the plates and the surface charge density. (d) What plate area would a $1.0$ F capacitor need with a gap of $1.0$ mm?
(a) Capacitance $A=(0.10)^2=1.0\times10^{-2}\ \text{m}^2$. $C=\dfrac{\varepsilon_0A}{d}=\dfrac{(8.85\times10^{-12})(1.0\times10^{-2})}{2.0\times10^{-3}}\approx4.4\times10^{-11}$ F $=44$ pF.
(b) Charge $Q=CV=(4.4\times10^{-11})(100)=4.4\times10^{-9}$ C: $+4.4$ nC on the plate joined to the $+$ terminal, $-4.4$ nC on the other.
(c) Field $E=\dfrac{V}{d}=\dfrac{100}{2.0\times10^{-3}}=5.0\times10^4$ V/m. $\sigma=\dfrac QA=4.4\times10^{-7}\ \text{C/m}^2$, and indeed $\sigma/\varepsilon_0=5.0\times10^4$ V/m ✓.
(d) One farad $A=\dfrac{Cd}{\varepsilon_0}=\dfrac{(1.0)(1.0\times10^{-3})}{8.85\times10^{-12}}\approx1.1\times10^{8}\ \text{m}^2$ — a square more than $10$ km on a side. That is why a farad is "enormous"; real high-capacitance devices use huge areas rolled up and gaps of molecular size.
Check Interactive 2 starts with this capacitor (area $100\ \text{cm}^2$, gap $2.0$ mm, $100$ V). Pause it and set the slab thickness to $0$: it reads $C=44.3$ pF, $Q=4.43$ nC, $E=5.0\times10^4$ V/m.

The Effect of a Dielectric

Fill the space between the plates with a dielectric of dielectric constant $K$ while keeping the charge $\pm Q$ on the plates fixed. The bound charges on the faces of the dielectric reduce the field from $E_0=\sigma/\varepsilon_0$ to $E_0/K$, so the potential difference falls from $V_0$ to $V_0/K$ and the capacitance rises by the factor $K$. If the dielectric fills only a slab of thickness $t$ (parallel to the plates), the field is $E_0$ in the air gaps, of total thickness $d-t$, and $E_0/K$ in the slab, so $$V=E_0(d-t)+\frac{E_0}{K}\,t=\frac{Q}{\varepsilon_0A}\left(d-t+\frac tK\right).$$

📐 Result 2.12 — Parallel-plate capacitor with a dielectric

(i) Completely filled with a dielectric of dielectric constant $K$: $\quad C=\dfrac{K\varepsilon_0A}{d}=KC_0 .$

(ii) With a dielectric slab of thickness $t\lt d$ parallel to the plates (the rest air): $$C=\frac{\varepsilon_0A}{d-t+\dfrac tK}.$$ (iii) A metal slab of thickness $t$ ($K\to\infty$): $C=\dfrac{\varepsilon_0A}{d-t}$. The position of the slab between the plates does not matter in (ii) or (iii).

The dielectric constant can therefore be measured as $K=C/C_0$, the factor by which a material filling the gap raises the capacitance.

What else changes when a dielectric is inserted depends on whether the capacitor is still connected to its battery:

Dielectric fills the gap$C$$Q$$V$$E$ between plates
Battery disconnected (isolated)$KC_0$$Q_0$ (no path for charge)$V_0/K$$E_0/K$
Battery connected$KC_0$$KQ_0$ (battery supplies more)$V_0$ (fixed by battery)$E_0$ (since $E=V/d$)
✏️ Example 6 — Inserting a dielectric, two ways
The capacitor of Example 5 ($C_0=44$ pF, charged to $100$ V) is used again. (a) It is disconnected from the battery and a slab with $K=4.0$ is slid in to fill the gap completely. Find the new capacitance, charge, potential difference and field. (b) Repeat with the battery left connected. (c) With the battery connected, the full slab is replaced by one of the same material but only $1.0$ mm thick. Find the capacitance and the charge.
(a) Isolated $C=KC_0=4.0\times44.3=177$ pF $\approx1.8\times10^{-10}$ F. $Q$ cannot change: $Q=4.4$ nC. $V=\dfrac QC=\dfrac{V_0}{K}=25$ V. $E=\dfrac{V}{d}=\dfrac{25}{2.0\times10^{-3}}=1.25\times10^4$ V/m — a quarter of the original.
(b) Connected $C=177$ pF, $V=100$ V (the battery holds it), $Q=CV=(1.77\times10^{-10})(100)\approx1.8\times10^{-8}$ C $=18$ nC — four times as much; the battery pushed the extra charge on. $E=V/d=5.0\times10^4$ V/m, unchanged.
(c) Partial slab $d-t+\dfrac tK=2.0-1.0+\dfrac{1.0}{4.0}=1.25$ mm. $C=\dfrac{(8.85\times10^{-12})(1.0\times10^{-2})}{1.25\times10^{-3}}\approx7.1\times10^{-11}$ F $=71$ pF; $Q=CV\approx7.1$ nC.
Check In Interactive 2, set $K=4$, slab thickness $2.0$ mm, and toggle the battery: the readouts follow (a) and (b). With $t=1.0$ mm you get $70.8$ pF ✓.
⚠️ Common mistakes with conductors, dielectrics and capacitors

1. "No field inside a conductor" ≠ "no potential". The whole conductor is at one potential, which can be large.

2. Forgetting what is held fixed. Before using $C=Q/V$, ask: is the battery still connected ($V$ fixed) or not ($Q$ fixed)? Most capacitor problems turn on this one question.

3. Using $C=K\varepsilon_0A/d$ for a partly filled gap. A slab of thickness $t\lt d$ needs $C=\varepsilon_0A/(d-t+t/K)$.

4. Thinking capacitance depends on charge. $C$ is fixed by geometry and material; double $Q$ and $V$ doubles too.

5. Mixing units. Area in m², gap in m: $10\ \text{cm}\times10\ \text{cm}=1.0\times10^{-2}\ \text{m}^2$, not $100$.

Interactive 2 — Parallel-Plate Capacitor Lab

Change the plate area, the gap and the dielectric slab, with the battery connected ($V$ fixed) or disconnected ($Q$ fixed at its value when you open the switch). The $\pm$ signs on the plates show the charge and the lines in the gap the field (spaced closer where it is stronger). The strip on the right plots the potential across the gap: a steady fall in air, a gentler one through the slab. Press Play to slide the slab in and out.

1002.00.04.0
capacitance $C$—
charge $Q$—
potential difference $V$—
$E$ in air; in slab—
energy stored $\tfrac12CV^2$—
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — spherical and cylindrical capacitors

Not required by the CBSE syllabus. Capacitance is listed by JEE Main, NEET and JEE Advanced, but each names only the parallel-plate capacitor; spherical and cylindrical capacitors are standard applications that no syllabus names.

The method of the parallel-plate derivation — find $E$ from Gauss's law, integrate to get $V$, divide $Q$ by $V$ — works for any symmetric pair of conductors (uses integration).

Spherical capacitor: concentric spheres of radii $a\lt b$, charges $+Q$ (inner) and $-Q$. Between them $E=kQ/r^2$, so $V=kQ\left(\dfrac1a-\dfrac1b\right)$ and $$C=4\pi\varepsilon_0\,\frac{ab}{b-a}.$$ As $b\to\infty$ this becomes the isolated sphere, $4\pi\varepsilon_0a$; when $b-a=d\ll a$ it becomes $\varepsilon_0(4\pi a^2)/d$, a parallel-plate capacitor.

Cylindrical capacitor (a coaxial cable): radii $a\lt b$, length $L$. $E=\dfrac{\lambda}{2\pi\varepsilon_0r}$ with $\lambda=Q/L$, so $V=\dfrac{\lambda}{2\pi\varepsilon_0}\ln\dfrac ba$ and $$C=\frac{2\pi\varepsilon_0L}{\ln(b/a)} .$$ A coaxial cable with $b/a=3.5$ and a polythene filling ($K=2.3$) has about $100$ pF per metre. These formulas are on the JEE syllabus but not in the board course.

Exercises

I
Conductors in electrostatics. $k=9.0\times10^9\ \text{N m}^2\,\text{C}^{-2}$, $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$.
  • 1 Explain why, in electrostatics, the field inside a conductor is zero and any excess charge lies on its surface.
  • 2 A metal sphere of radius $15$ cm carries $30$ nC. Find the surface charge density, the field just outside, and the potential of the sphere. What are the field and potential at its centre?
  • 3 At a certain point on the surface of a charged conductor the surface charge density is $2.0\ \mu\text{C/m}^2$. Find the field just outside it.
  • 4 Metal spheres of radii $3.0$ cm and $9.0$ cm, far apart, are joined by a thin wire and share $24$ nC. Find the charge on each, their common potential and the ratio of their surface charge densities.
  • 5 Why is a person inside a car (or an aircraft) safe when it is struck by lightning? Would the person be equally safe in a car made of fibreglass?
  • 6 A point charge $+q$ is placed at the centre of a neutral metal shell. What charges appear on the inner and outer surfaces? If the charge is moved off-centre (still inside), which of these charge distributions changes, and does the field outside change?
  • 7 Show that the field is zero everywhere in an empty cavity inside a conductor, even when the conductor is placed in an external field.
  • 8 Modelling. On a fine day the field at the Earth's surface is about $130$ V/m, pointing downwards. Find the surface charge density on the ground and estimate the total charge on the Earth (surface area $5.1\times10^{14}\ \text{m}^2$).
Check: two spheres joined by a wire
$q_1$, $q_2$—
common potential—
$\sigma_1/\sigma_2$; surface fields—
1 If there were a field inside, the free electrons would feel a force and move; they keep moving until their induced charges cancel the field, so in equilibrium $E=0$. Then Gauss's law for any small closed surface inside the material gives zero enclosed charge: excess charge can only sit on the surface.
2 $\sigma=\dfrac{3.0\times10^{-8}}{4\pi(0.15)^2}\approx1.1\times10^{-7}\ \text{C/m}^2$; $E=\dfrac{kQ}{R^2}=\dfrac{270}{0.0225}=1.2\times10^4$ N/C; $V=\dfrac{270}{0.15}=1.8\times10^3$ V. Centre: $E=0$, $V=1.8\times10^3$ V.
3 $E=\dfrac{\sigma}{\varepsilon_0}=\dfrac{2.0\times10^{-6}}{8.85\times10^{-12}}\approx2.3\times10^{5}$ N/C, normal to the surface.
4 $q\propto R$: $q_1=24\times\dfrac{3}{12}=6.0$ nC, $q_2=18$ nC. $V=\dfrac{(9.0\times10^9)(6.0\times10^{-9})}{0.030}=1.8\times10^3$ V. $\dfrac{\sigma_1}{\sigma_2}=\dfrac{R_2}{R_1}=3$.
5 The metal body is a closed conductor: the charge stays on its outer surface and the field inside (the cavity) stays zero (electrostatic shielding); the current flows round the skin to the ground. Fibreglass is an insulator and gives no such shielding.
6 Inner surface $-q$, outer surface $+q$. Moving the charge off-centre changes the distribution on the inner surface (it crowds towards the charge), but the outer surface charge stays uniform — the metal screens the outside from details inside — so the field outside is unchanged: $kq/r^2$ from the centre.
7 The whole conductor, including the cavity wall, is at one potential. If there were a field in the empty cavity, a field line would have to start and end on the wall; going along it, $V$ would fall, so its two ends would be at different potentials — impossible. So $E=0$ in the cavity. (With no charge in the cavity, the wall also carries no charge.)
8 Just outside a conductor $E=\sigma/\varepsilon_0$: $|\sigma|=(8.85\times10^{-12})(130)\approx1.2\times10^{-9}\ \text{C/m}^2$, negative (the field points into the ground). $Q\approx(1.15\times10^{-9})(5.1\times10^{14})\approx-5.9\times10^{5}$ C — large, but spread over the whole planet.
II
Dielectrics and polarisation.
  • 1 Distinguish between polar and non-polar molecules, with two examples of each. How does each kind respond to an external field?
  • 2 A dielectric slab with $K=3.0$ is placed in a uniform field of $6.0\times10^4$ V/m normal to its faces. Find the field inside the slab and the bound surface charge density.
  • 3 When a slab is placed in a field of $1.0\times10^4$ V/m (normal to its faces) the field inside is $4.0\times10^3$ V/m. Find its dielectric constant.
  • 4 Explain, using bound charges, why a dielectric placed in a field weakens the field inside it but does not cancel it.
  • 5 A material has electric susceptibility $\chi_e=4.4\times10^{-11}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$. Find its dielectric constant.
  • 6 Why is a metal said to have an infinite dielectric constant?
1 Non-polar ($\mathrm{O_2}$, $\mathrm{CO_2}$, $\mathrm{CH_4}$): the centres of positive and negative charge coincide; a field induces a small dipole along the field. Polar ($\mathrm{H_2O}$, $\mathrm{HCl}$): permanent dipoles, randomly oriented; a field exerts a torque and partly aligns them (and also induces a little extra moment).
2 $E=\dfrac{6.0\times10^4}{3.0}=2.0\times10^4$ V/m. $\sigma_p=P=\varepsilon_0(E_0-E)=(8.85\times10^{-12})(4.0\times10^4)\approx3.5\times10^{-7}\ \text{C/m}^2$.
3 $K=\dfrac{E_0}{E}=\dfrac{1.0\times10^4}{4.0\times10^3}=2.5$.
4 The aligned dipoles leave a layer of negative bound charge on the face where the field enters and positive on the face where it leaves; their field inside points against $E_0$. It only partly cancels $E_0$ because the polarisation is itself caused by the field inside: if the field were cancelled, there would be nothing left to polarise the molecules.
5 $K=1+\dfrac{\chi_e}{\varepsilon_0}=1+\dfrac{4.4\times10^{-11}}{8.85\times10^{-12}}\approx1+4.97\approx6.0$.
6 Inside a metal $E=0$ for any applied $E_0$; $E=E_0/K$ then requires $K\to\infty$. Equivalently, a metal slab of thickness $t$ changes the capacitance exactly as a dielectric slab with $K\to\infty$ would.
III
Capacitance and the parallel-plate capacitor (air between the plates unless stated).
  • 1 A capacitor stores $6.0\ \mu\text{C}$ when the potential difference across it is $12$ V. Find its capacitance.
  • 2 Find the capacitance of square plates of side $20$ cm placed $1.0$ mm apart.
  • 3 What gap is needed between plates of area $0.20\ \text{m}^2$ for a capacitance of $1.0$ nF?
  • 4 Treating the Earth as an isolated conducting sphere of radius $6.4\times10^6$ m, find its capacitance.
  • 5 What is the radius of an isolated sphere whose capacitance is $1.0$ pF?
  • 6 A parallel-plate capacitor has $C=50$ pF. Its plate area is doubled and the gap halved. Find the new capacitance.
  • 7 A capacitor with a $0.50$ mm air gap is connected to $100$ V. Find the field in the gap. Air breaks down at about $3.0\times10^6$ V/m: what is the greatest voltage this capacitor can take?
  • 8 Show from $C=\varepsilon_0A/d$ that $\varepsilon_0$ may be measured in F/m, and check that $1\ \text{F/m}=1\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$.
1 $C=\dfrac QV=\dfrac{6.0\times10^{-6}}{12}=5.0\times10^{-7}$ F $=0.50\ \mu\text{F}$.
2 $C=\dfrac{(8.85\times10^{-12})(0.040)}{1.0\times10^{-3}}\approx3.5\times10^{-10}$ F $=350$ pF.
3 $d=\dfrac{\varepsilon_0A}{C}=\dfrac{(8.85\times10^{-12})(0.20)}{1.0\times10^{-9}}\approx1.8\times10^{-3}$ m $=1.8$ mm.
4 $C=4\pi\varepsilon_0R=\dfrac Rk=\dfrac{6.4\times10^6}{9.0\times10^9}\approx7.1\times10^{-4}$ F $\approx710\ \mu\text{F}$ — the whole planet has less capacitance than a large electrolytic capacitor.
5 $R=kC=(9.0\times10^9)(1.0\times10^{-12})=9.0\times10^{-3}$ m $=9.0$ mm.
6 $C\propto A/d$: $C'=50\times2\times2=200$ pF.
7 $E=\dfrac{100}{5.0\times10^{-4}}=2.0\times10^5$ V/m. $V_\text{max}=E_\text{max}d=(3.0\times10^6)(5.0\times10^{-4})=1.5\times10^3$ V.
8 $\varepsilon_0=\dfrac{Cd}{A}$ has unit $\dfrac{\text{F m}}{\text{m}^2}=\text{F/m}$. With $1\ \text{F}=1\ \text{C/V}=1\ \text{C}^2/\text{J}=1\ \text{C}^2/(\text{N m})$: $1\ \text{F/m}=1\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$ ✓.
IV
Dielectrics in capacitors. $\varepsilon_0=8.85\times10^{-12}$ F/m.
  • 1 An air capacitor of $20$ pF is completely filled with a dielectric of $K=5.0$. Find its new capacitance.
  • 2 An air capacitor is charged to $200$ V and disconnected. A dielectric with $K=4.0$ is then inserted to fill the gap. How do $Q$, $V$ and $E$ change? Give the new $V$.
  • 3 Repeat part 2 with the battery left connected.
  • 4 Plates of area $0.010\ \text{m}^2$ are $4.0$ mm apart. A slab $2.0$ mm thick with $K=2.0$ is placed between them, parallel to the plates. Find the capacitance.
  • 5 In part 4, the dielectric slab is replaced by a metal slab $2.0$ mm thick. Find the capacitance. Does it matter where in the gap the slab is placed?
  • 6 Modelling. A key on a capacitive keyboard has plates of area $1.0\ \text{cm}^2$ separated by $0.60$ mm of air. Pressing the key reduces the gap to $0.15$ mm. Find the capacitance before and after, and the factor by which it changes.
  • 7 Modelling. A $10$ nF capacitor must withstand $1.0$ kV. It is to be made with a dielectric of $K=3.0$ that breaks down at $2.0\times10^7$ V/m. Find the least possible gap and the plate area needed.
1 $C=KC_0=5.0\times20=100$ pF.
2 Isolated: $Q$ unchanged; $C$ becomes $4C_0$, so $V=\dfrac{200}{4.0}=50$ V; $E=V/d$ falls to a quarter.
3 Connected: $V$ stays $200$ V and $E=V/d$ is unchanged; $C$ and so $Q=CV$ both become four times as large.
4 $d-t+\dfrac tK=4.0-2.0+1.0=3.0$ mm. $C=\dfrac{(8.85\times10^{-12})(0.010)}{3.0\times10^{-3}}\approx2.95\times10^{-11}$ F $\approx29.5$ pF.
5 $C=\dfrac{\varepsilon_0A}{d-t}=\dfrac{8.85\times10^{-14}}{2.0\times10^{-3}}\approx4.4\times10^{-11}$ F $\approx44$ pF. Position does not matter: only the total thickness of air, $d-t$, enters.
6 Before: $C=\dfrac{(8.85\times10^{-12})(1.0\times10^{-4})}{6.0\times10^{-4}}\approx1.5$ pF. After: $\dfrac{8.85\times10^{-16}}{1.5\times10^{-4}}\approx5.9$ pF. $C\propto1/d$, so it rises by a factor of $4$ — easily detected by the keyboard's circuit.
7 $d_\text{min}=\dfrac{V}{E_\text{max}}=\dfrac{1.0\times10^3}{2.0\times10^7}=5.0\times10^{-5}$ m $=50\ \mu$m. $A=\dfrac{Cd}{K\varepsilon_0}=\dfrac{(1.0\times10^{-8})(5.0\times10^{-5})}{3.0\times8.85\times10^{-12}}\approx1.9\times10^{-2}\ \text{m}^2$ — about $14$ cm square, usually rolled up into a small cylinder. (In practice a safety margin is added to the gap.)

Original problems in the style of the CBSE board paper, JEE Main, JEE Advanced and NEET. Try each one against the clock (about 2 minutes for an MCQ, 5 minutes for a long answer) before opening the solution.

🎯 Exam corner — conductors, dielectrics and capacitors BoardJEE MainJEE AdvancedNEET

E1. Board · 1 mark Assertion (A): A person sitting inside a closed metal car is safe from a lightning strike on the car. Reason (R): The electric field inside an empty cavity in a conductor is zero. Choose: (A) both A and R true and R is the correct explanation of A; (B) both true but R is not the correct explanation; (C) A true, R false; (D) A false, R true.

Concept Electrostatic shielding: charges on a closed conductor stay on its outer surface and the field in the cavity is zero.
Answer (A). Exam tip: the car is not perfectly closed and the charge is flowing, so the shielding is not exact — but the principle is the one stated in R.

E2. Board · 3 marks A parallel-plate capacitor has plates of area $A$ separated by $d$. A dielectric slab of thickness $t\lt d$ and dielectric constant $K$ is placed between the plates, parallel to them. Derive an expression for the capacitance. Evaluate it for $A=0.020\ \text{m}^2$, $d=3.0$ mm, $t=2.0$ mm and $K=4.0$.

Fields Charge $\pm Q$, density $\sigma=Q/A$. In the air: $E_0=\dfrac{\sigma}{\varepsilon_0}$. In the slab the bound charges reduce it to $E=\dfrac{E_0}{K}$.
Potential $V=E_0(d-t)+\dfrac{E_0}{K}t=\dfrac{Q}{\varepsilon_0A}\left(d-t+\dfrac tK\right)$.
Capacitance $C=\dfrac QV=\dfrac{\varepsilon_0A}{d-t+t/K}$. Checks: $t=0$ gives $\varepsilon_0A/d$; $t=d$ gives $K\varepsilon_0A/d$.
Number $d-t+t/K=3.0-2.0+0.50=1.5$ mm. $C=\dfrac{(8.85\times10^{-12})(0.020)}{1.5\times10^{-3}}\approx1.2\times10^{-10}$ F $\approx118$ pF. Exam tip: the two limiting-case checks take ten seconds and often earn the last mark.

E3. JEE Main (Numerical value.) An air-filled parallel-plate capacitor is charged to $60$ V and then disconnected from the battery. A slab of dielectric constant $3$ is inserted so that it fills the space between the plates completely. Find the new potential difference, in volts.

Concept Isolated: $Q$ is fixed; $C\to KC$, so $V=Q/C\to V/K$.
Answer $V=\dfrac{60}{3}=\mathbf{20}$ V. Exam tip: write "Q fixed" or "V fixed" before anything else.

E4. NEET Two metal spheres of radii $R$ and $2R$, far apart, are joined by a thin wire and charged. The ratio of the surface charge density on the smaller sphere to that on the larger one is (A) $1/2$ (B) $2$ (C) $4$ (D) $1$.

Concept Same potential: $q_1/R_1=q_2/R_2$, so $q\propto R$ and $\sigma=\dfrac{q}{4\pi R^2}\propto\dfrac1R$.
Answer (B) $\dfrac{\sigma_1}{\sigma_2}=\dfrac{2R}{R}=2$. Exam tip: the charges are in the ratio $1:2$ but the densities in the ratio $2:1$ — the examiners hope you will mix them up.

E5. JEE Advanced (One or more correct.) An air-filled parallel-plate capacitor of capacitance $C_0$ and gap $d$ stays connected to a battery of emf $V_0$. A dielectric slab of thickness $d/2$ and dielectric constant $2$ is inserted parallel to the plates. Before insertion the field in the gap is $E_0$ and the charge $Q_0$. After insertion: (A) the capacitance is $\tfrac43C_0$; (B) the field in the air part is $\tfrac43E_0$; (C) the field inside the slab is $\tfrac23E_0$; (D) the charge on the plates is $2Q_0$.

Capacitance $d-t+t/K=d-\tfrac d2+\tfrac d4=\tfrac34d$, so $C=\tfrac43C_0$: (A) ✓.
Charge $V_0$ fixed: $Q=CV_0=\tfrac43Q_0$, so (D) ✗.
Fields In the air $E=\sigma/\varepsilon_0\propto Q$: $E_\text{air}=\tfrac43E_0$ (B) ✓. In the slab $E=E_\text{air}/K=\tfrac23E_0$ (C) ✓. Check: $V=\tfrac43E_0\cdot\tfrac d2+\tfrac23E_0\cdot\tfrac d2=E_0d=V_0$ ✓.
Answer (A), (B), (C). Exam tip: always finish with the check $\sum E_i t_i=V$; it catches most sign and factor slips.

E6. Board · Case study A thick, neutral, hollow metal sphere has inner radius $9.0$ cm and outer radius $10$ cm. A charge of $+2.0$ nC is held at its centre on an insulating rod. (a) State the charges on the inner and outer surfaces of the sphere and find the field at $5.0$ cm and at $15$ cm from the centre. (2 marks) (b) What is the field within the metal? (1 mark) (c) The sphere is now earthed. What is the field at $15$ cm? (1 mark)

(a) Inner surface $-2.0$ nC, outer surface $+2.0$ nC. $E(5.0\ \text{cm})=\dfrac{(9.0\times10^9)(2.0\times10^{-9})}{(0.050)^2}=7.2\times10^3$ N/C; $E(15\ \text{cm})=\dfrac{18}{(0.15)^2}=8.0\times10^2$ N/C, both radially outwards.
(b) Zero — it is inside a conductor in equilibrium.
(c) Earthing removes the $+2.0$ nC from the outer surface; the enclosed charge at $15$ cm is then $+2.0-2.0=0$, so $E=0$. Exam tip: earthing a shell removes only the charge on its outer surface; the inner surface charge is held by the charge inside.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. Can you measure the dielectric constant of ordinary printer paper with two sheets of kitchen foil and a multimeter — and how do you tell a real trend from experimental error?

Why it matters. Measuring the dielectric constant of materials is routine work in materials laboratories: it decides how food heats in a microwave oven, how far a radar signal penetrates soil or ice, and which plastics can be used inside high-frequency circuit boards. Capacitive sensors that read the moisture in grain silos in Canada or soil in Indian farms work because water's $K\approx80$ is so much larger than that of dry material.

What to do.

  1. The set-up: two foil sheets $20\ \text{cm}\times20$ cm, separated by $n$ sheets of paper each $0.10$ mm thick, pressed flat under a heavy book; the capacitance is read on a multimeter. Sample results: $n=1$: $9.6$ nF; $n=2$: $5.1$ nF; $n=3$: $3.5$ nF; $n=4$: $2.6$ nF; $n=6$: $1.8$ nF.
  2. Theory says $C=K\varepsilon_0A\cdot\dfrac1d$. Plot $C$ against $1/d$. Is it a straight line through the origin?
  3. Find the best slope of a line through the origin (by eye, or $\sum xy/\sum x^2$ with $x=1/d$, $y=C$) and hence $K$.
  4. The point for $n=1$ falls below the line. Suggest a physical reason (think about the foil not lying perfectly flat), and say whether you would include it.

Tools: Spreadsheet or graph paper; optionally real foil, paper and a multimeter with a capacitance range.

Hand in: The table of $d$, $1/d$ and $C$; the graph; your value of $K$ with and without the $n=1$ point; and a short discussion of errors.

Going further. Repeat the analysis with a sheet of plastic (overhead-projector film) and compare. Why do measured dielectric constants depend on the frequency of the meter's test signal, especially for materials that contain water?

Instructor note: $\varepsilon_0A=3.54\times10^{-13}$ F m. Products $Cd$ are $0.96,\ 1.02,\ 1.05,\ 1.04,\ 1.08\times10^{-12}$ F m. The least-squares slope through the origin is about $0.98\times10^{-12}$ F m (dominated by the $n=1$ point), giving $K\approx2.8$; without $n=1$ the mean $Cd\approx1.05\times10^{-12}$ gives $K\approx3.0$. Thin air gaps between foil and paper add a series air layer whose effect is largest when the paper is thin, so $n=1$ reads low. Paper is typically $K\approx2$–$4$.

🧪 Self-check quiz

0/5 correct

Q1. A hollow metal sphere carries a charge $Q$. The field at a point inside the hollow is

Q2. A charged parallel-plate capacitor is disconnected from its battery, and a dielectric slab is then pushed in to fill the gap. The potential difference between the plates

Q3. One farad is equal to

Q4. The plate area of an air-filled parallel-plate capacitor is doubled and the gap is also doubled. Its capacitance

Q5. A metal slab of thickness $d/2$ is placed between the plates of an air-filled capacitor of gap $d$, parallel to them. The capacitance becomes

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