Chapter 1 · Electric Charges and Fields

1.1 Electric Charge, Conductors and Coulomb's Law

Charge comes in two kinds, is conserved and comes in whole-number multiples of $e$. Two point charges push or pull with $F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{|q_1q_2|}{r^2}$, and with many charges the forces simply add as vectors.

Peel a strip of sticky tape off a desk and it reaches for your hand. Pull off a woollen sweater in winter and you hear crackles; touch a car door after a dry drive and you feel a spark. A laser printer uses the same effect on purpose: a drum is charged in the pattern of your page, toner powder sticks only where the charge is, and the image is pressed onto paper. Behind all of these is one property of matter — electric charge — and one force law telling us how charges push and pull on each other.

This section sets up the rules of the game for the whole of electrostatics. First we list what charge is and how it behaves: two kinds, conserved, and counted in whole numbers of the electron's charge. Then we see how charge moves easily through some materials and not others, which lets us charge a metal sphere without touching it. Finally we meet Coulomb's law, the inverse-square law for the force between two point charges, and the principle of superposition, which lets us find the force from any number of charges by adding vectors. Every later idea in this chapter — fields, field lines, dipoles, flux and Gauss's law — is built on these two laws.

Electric Charge and Its Properties

Rub a glass rod with silk and a plastic comb with dry hair. The two rubbed objects attract each other, while two rubbed glass rods repel, and so do two rubbed combs. Experiments like this, done carefully with many materials, show that there are exactly two kinds of charge. Benjamin Franklin named them positive (the kind on the glass rod) and negative (the kind on the comb). The rule is simple: like charges repel, unlike charges attract.

Today we know where the two kinds come from. Ordinary matter is made of atoms with a tiny positive nucleus (protons and neutrons) surrounded by negative electrons. A proton carries charge $+e$ and an electron carries $-e$, where $$e=1.6\times10^{-19}\ \text{C}.$$ A neutral atom has equal numbers of protons and electrons. Rubbing does not create charge: it transfers a few electrons from one surface to the other. The object that gains electrons becomes negative; the object that loses them is left positive by exactly the same amount.

📘 Definition 1.1 — Electric charge and the coulomb
Electric charge is the property of matter that makes it exert and feel electric forces. It is a scalar that can be positive or negative. A body is neutral when its total charge is zero, positively charged when it has fewer electrons than protons, and negatively charged when it has more. The SI unit of charge is the coulomb (C); in terms of base units $1\ \text{C}=1\ \text{A s}$ (the charge carried past a point in one second by a steady current of one ampere). Useful sub-units: $1\ \mu\text{C}=10^{-6}\ \text{C}$, $1\ \text{nC}=10^{-9}\ \text{C}$.
📐 Principle 1.1 — The three basic properties of charge

1. Additivity. Charges add like ordinary signed numbers: a system containing charges $q_1,q_2,\dots,q_n$ has total charge $Q=q_1+q_2+\cdots+q_n$ (signs included). Mass is also additive, but mass is never negative; charge can cancel.

2. Conservation. The total charge of an isolated system never changes. Charge can be transferred from one body to another, and pairs of opposite charges can be created or destroyed together (for example, a high-energy photon can become an electron and a positron), but the net charge of the isolated system stays the same.

3. Quantisation. Every free charge that has ever been observed is a whole-number multiple of the elementary charge: $$q=ne,\qquad n=0,\pm1,\pm2,\pm3,\dots$$

Why do we not notice quantisation in daily life? Because $e$ is so small. A charge of only $1\ \mu\text{C}$ already contains about $6\times10^{12}$ elementary charges, so adding or removing one electron changes the charge by less than a millionth of a millionth. At the scale of combs and balloons, charge behaves as if it could take any value; at the scale of atoms and electrons, it comes in steps of $e$. (Quarks inside protons carry $\pm\tfrac13e$ or $\pm\tfrac23e$, but they are never found free, so the rule $q=ne$ holds for every charge we can isolate and measure.)

✏️ Example 1 — Counting electrons on a charged strip
A polythene strip rubbed with wool acquires a charge of $-4.0\ \text{nC}$. (a) How many electrons were transferred, and in which direction? (b) By how much did the mass of the strip change? (c) Could any object carry a charge of $2.4\times10^{-19}\ \text{C}$? Take $m_e=9.1\times10^{-31}$ kg.
Concept Charge is quantised, $q=ne$, and a negative charge means extra electrons.
(a) Count $$n=\frac{|q|}{e}=\frac{4.0\times10^{-9}\ \text{C}}{1.6\times10^{-19}\ \text{C}}=2.5\times10^{10}.$$ The strip is negative, so $2.5\times10^{10}$ electrons moved from the wool to the polythene. The wool is left with $+4.0$ nC (conservation).
(b) Mass $\Delta m=n\,m_e=(2.5\times10^{10})(9.1\times10^{-31}\ \text{kg})\approx2.3\times10^{-20}\ \text{kg}$ — an increase far too small to weigh.
(c) Quantisation $\dfrac{2.4\times10^{-19}}{1.6\times10^{-19}}=1.5$, not a whole number. No: a free charge of $1.5e$ is impossible.
✏️ Example 2 — Sharing charge between identical spheres
Three identical small metal spheres $A$, $B$ and $C$ carry charges $+8.0\ \mu\text{C}$, $-2.0\ \mu\text{C}$ and $0$. Sphere $A$ is touched to $B$ and separated; then $B$ is touched to $C$ and separated. Find the final charge on each sphere and check that charge is conserved.
Idea When two identical conductors touch, charge flows until they are in the same electrical state; by symmetry each ends with half of their combined charge.
A touches B Combined charge $(+8.0)+(-2.0)=+6.0\ \mu\text{C}$, so $A$ and $B$ each carry $+3.0\ \mu\text{C}$.
B touches C Combined charge $(+3.0)+0=+3.0\ \mu\text{C}$, so $B$ and $C$ each carry $+1.5\ \mu\text{C}$.
Check Final: $A=+3.0$, $B=+1.5$, $C=+1.5\ \mu\text{C}$. Total $=+6.0\ \mu\text{C}$, the same as the initial total $8.0-2.0+0=6.0\ \mu\text{C}$. ✓ (The halving rule works only for identical spheres; for unequal spheres the larger one takes more.)

Conductors, Insulators and Charging by Induction

Hold a metal rod in your hand and rub it: it does not seem to charge. Rub a plastic rod and it does. The difference is whether charge can move through the material.

📘 Definition 1.2 — Conductors, insulators and earthing
A conductor is a material in which some charges are free to move through the bulk — metals (free electrons), the human body, tap water and the Earth. An insulator is a material in which charges stay where they are placed — glass, dry wood, plastic, rubber, porcelain. Charge put on an insulator stays near that spot; charge put on a conductor spreads over its whole outer surface at once.
Earthing (grounding) means connecting a conductor to the Earth by a conducting path. The Earth is so large that it can give or take any amount of charge without its own state changing noticeably, so an earthed conductor shares its excess charge with the Earth and is left neutral unless other charges nearby hold charge on it.

This explains the metal rod: the charge you rub onto it flows through the rod and your body into the Earth. Give the rod an insulating handle and it charges perfectly well. (Semiconductors such as silicon lie in between; they are the subject of Chapter 14.)

Free electrons in a conductor also explain a striking fact: a charged rod attracts a neutral metal sphere. Bring a negative rod near the sphere and its free electrons are pushed to the far side, leaving the near side positive. The sphere is still neutral overall, but its positive charge is closer to the rod than its negative charge, and because electric force weakens with distance the attraction wins. This separation of charge by a nearby charged body, without contact, is called induction. Even in an insulator such as paper the molecules become slightly lopsided (polarised), which is why a rubbed comb picks up small bits of paper.

🧭 Problem-solving strategy — charging a conductor by induction
  1. Bring the charged rod near (not touching) the insulated conductor. Free charges in the conductor separate: opposite charge gathers on the near side, like charge on the far side.
  2. Earth the conductor (touch it with a finger or a wire) while the rod is still near. The like charge on the far side is pushed away into the Earth; the opposite charge stays, held by the attraction of the rod.
  3. Remove the earth connection first, with the rod still in place, so the charge cannot flow back.
  4. Then remove the rod. The remaining charge spreads over the whole surface. The conductor now carries charge opposite to the rod's, and the rod has lost none of its own charge.
⚠️ Common mistakes about charging

1. Protons do not move. In solids only electrons are transferred. "A rod becomes positive" always means it lost electrons, never that it gained protons.

2. Order matters in induction. If you remove the rod before breaking the earth connection, electrons flow back from the Earth and the conductor ends up neutral.

3. Attraction is not proof of charge. A charged rod attracts both an oppositely charged body and a neutral body (by induction). Only repulsion proves that a body is charged, and with the same sign.

Coulomb's Law

In 1785 Charles-Augustin de Coulomb measured the force between small charged spheres with a delicate torsion balance. He found that the force acts along the line joining the charges, is proportional to each charge, and falls off as the inverse square of the distance — the same form as Newton's law of gravitation, but with charges in place of masses. The law applies to point charges: charged bodies whose sizes are much smaller than the distance between them.

📐 Law 1.2 — Coulomb's law

Two stationary point charges $q_1$ and $q_2$ a distance $r$ apart in vacuum exert forces on each other of magnitude $$F=k\,\frac{|q_1q_2|}{r^2},\qquad k=\frac{1}{4\pi\varepsilon_0}\approx9.0\times10^{9}\ \text{N m}^2\,\text{C}^{-2},$$ directed along the line joining them: repulsive if the charges have the same sign, attractive if opposite. Here $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$ is the permittivity of free space.

Vector form. Let $\mathbf r_{21}=\mathbf r_2-\mathbf r_1$ point from $q_1$ to $q_2$ and $\hat{\mathbf r}_{21}=\mathbf r_{21}/r$. The force on $q_2$ due to $q_1$ is $$\mathbf F_{21}=\frac{1}{4\pi\varepsilon_0}\,\frac{q_1q_2}{r^2}\,\hat{\mathbf r}_{21},$$ with the signs of $q_1,q_2$ included: $q_1q_2\gt0$ gives a force along $\hat{\mathbf r}_{21}$ (away from $q_1$), $q_1q_2\lt0$ a force towards $q_1$. Swapping the labels reverses $\hat{\mathbf r}$, so $\mathbf F_{12}=-\mathbf F_{21}$: Coulomb forces obey Newton's third law.

In a medium. If the charges sit in a large uniform insulating medium (oil, water, glass), the force is smaller by the relative permittivity (dielectric constant) $\varepsilon_r\ge1$ of the medium: $F=\dfrac{1}{4\pi\varepsilon_0\varepsilon_r}\dfrac{|q_1q_2|}{r^2}$. Air has $\varepsilon_r\approx1.0006$, so air and vacuum are treated alike.

The constant $k$ is not a rounded oddity: the SI defines things so that $k=\dfrac{1}{4\pi\varepsilon_0}$, and the factor $4\pi$ is placed here so that it disappears from Gauss's law later in this chapter. In numerical work use $k=9.0\times10^9\ \text{N m}^2\,\text{C}^{-2}$. Notice how enormous it is: two charges of $1$ C, a metre apart, would push on each other with $9\times10^9$ N — the weight of about a million cars. A coulomb is a huge amount of static charge; everyday static charges are microcoulombs or nanocoulombs.

✏️ Example 3 — Force between two charges, in air and in water
Point charges $q_1=+2.0\ \mu\text{C}$ and $q_2=-3.0\ \mu\text{C}$ are $10$ cm apart. (a) Find the force between them in air. (b) Find the force if both are in water ($\varepsilon_r=80$) at the same separation. (c) How far apart would they have to be in air to feel the force of part (b)?
Setup Convert to SI first: $q_1=2.0\times10^{-6}$ C, $|q_2|=3.0\times10^{-6}$ C, $r=0.10$ m. Opposite signs, so the force is attractive.
(a) Air $$F=k\frac{|q_1q_2|}{r^2}=\frac{(9.0\times10^9)(2.0\times10^{-6})(3.0\times10^{-6})}{(0.10)^2}=\frac{5.4\times10^{-2}}{1.0\times10^{-2}}=5.4\ \text{N},$$ attractive — each charge is pulled towards the other.
(b) Water $F_w=\dfrac{F}{\varepsilon_r}=\dfrac{5.4\ \text{N}}{80}\approx6.8\times10^{-2}\ \text{N}$ (still attractive). Water's large $\varepsilon_r$ is why salts such as NaCl fall apart into ions when they dissolve: the attraction holding the ions together is weakened eighty-fold.
(c) Equal force We need $\dfrac{k|q_1q_2|}{r'^2}=\dfrac{k|q_1q_2|}{\varepsilon_r r^2}$, so $r'=r\sqrt{\varepsilon_r}=(0.10\ \text{m})\sqrt{80}\approx0.89$ m.
Check Set $q_1=+2$, $q_2=-3$ and $r=10$ cm in Interactive 1: the readout shows $5.4$ N in vacuum and $0.0675$ N in water. ✓
✏️ Example 4 — Electric force versus gravity inside a hydrogen atom
In a hydrogen atom the electron and proton are about $r=0.53\ \text{Å}=5.3\times10^{-11}$ m apart. Compare the electric force between them with the gravitational force. Take $G=6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$, $m_e=9.1\times10^{-31}$ kg, $m_p=1.67\times10^{-27}$ kg.
Electric $$F_e=\frac{ke^2}{r^2}=\frac{(9.0\times10^9)(1.6\times10^{-19})^2}{(5.3\times10^{-11})^2}\approx8.2\times10^{-8}\ \text{N}.$$
Gravity $$F_g=\frac{Gm_em_p}{r^2}=\frac{(6.67\times10^{-11})(9.1\times10^{-31})(1.67\times10^{-27})}{(5.3\times10^{-11})^2}\approx3.6\times10^{-47}\ \text{N}.$$
Ratio $\dfrac{F_e}{F_g}=\dfrac{ke^2}{Gm_em_p}\approx2.3\times10^{39}$. The distance cancels, so the ratio is the same at every separation.
Meaning Gravity is utterly negligible between charged particles; atoms, molecules and the solidity of the chair you sit on are electrical. Gravity dominates for planets only because large bodies are almost perfectly neutral, so their huge electric forces cancel out.
Interactive 1 — Coulomb's Law Lab

Charge $q_1$ is fixed at the left end of the track; drag $q_2$ (or use the $r$ slider). The arrows are the forces on each charge — always equal in size and opposite in direction. The graph above the track plots $F$ against $r$ on the same horizontal scale: halve $r$ and the force quadruples. Change the medium and the whole curve drops by $\varepsilon_r$ (the dashed curve is vacuum).

+2−3
separation $r$—
force $F$ on each charge—
nature—
$F\,r^2$ (N m²) — constant—

Forces between Multiple Charges: Superposition

Coulomb's law is about two charges. Real problems — a charge near several others, an ion in a crystal, toner near a printer drum — involve many. Experiment answers the question of how they combine, and the answer is the simplest possible one.

📐 Principle 1.3 — The superposition principle

The force between any two charges is not affected by the presence of other charges. Hence, in a system of point charges $q_1,q_2,\dots,q_n$, the net force on $q_1$ is the vector sum of the forces exerted on it by each of the others, each calculated from Coulomb's law as if the rest were absent: $$\mathbf F_1=\mathbf F_{12}+\mathbf F_{13}+\cdots+\mathbf F_{1n}=\frac{q_1}{4\pi\varepsilon_0}\sum_{i=2}^{n}\frac{q_i}{r_{1i}^{2}}\,\hat{\mathbf r}_{1i},$$ where $r_{1i}$ is the distance from $q_i$ to $q_1$ and $\hat{\mathbf r}_{1i}$ is the unit vector pointing from $q_i$ towards $q_1$.

🧭 Problem-solving strategy — net force on one charge
  1. Draw the charges with coordinates. Mark the charge you want the force on.
  2. For each other charge, find the magnitude $F_i=k|q_iq_0|/r_i^2$ (no signs) and decide the direction from the signs: away from $q_i$ if like, towards $q_i$ if unlike. Draw each force as an arrow starting at $q_0$.
  3. Resolve each force into $x$- and $y$-components using the geometry (unit vectors, or $\cos$ and $\sin$ of known angles). Use symmetry to spot components that cancel.
  4. Add the components: $F_x=\sum F_{ix}$, $F_y=\sum F_{iy}$. Then $|\mathbf F|=\sqrt{F_x^2+F_y^2}$ and $\theta=\tan^{-1}(F_y/F_x)$, choosing the quadrant from the signs.
✏️ Example 5 — Net force at the corner of a right triangle
Charge $q_1=+3.0\ \mu\text{C}$ is at $A=(0,0)$ and $q_2=-4.0\ \mu\text{C}$ is at $B=(12\ \text{cm},0)$. A charge $q_0=+1.0\ \mu\text{C}$ is at $C=(0,9\ \text{cm})$. Find the net force on $q_0$ (magnitude and direction).
Geometry $AC=9$ cm, $AB=12$ cm, so $BC=\sqrt{9^2+12^2}=15$ cm (a 3–4–5 triangle). The unit vector from $C$ towards $B$ is $\dfrac{(12,-9)}{15}=(0.8,\,-0.6)$.
Force from $q_1$ $F_1=\dfrac{(9.0\times10^9)(3.0\times10^{-6})(1.0\times10^{-6})}{(0.09)^2}\approx3.33$ N. Like charges, so it pushes $q_0$ away from $A$: straight up, $\mathbf F_1=(0,\ 3.33)$ N.
Force from $q_2$ $F_2=\dfrac{(9.0\times10^9)(4.0\times10^{-6})(1.0\times10^{-6})}{(0.15)^2}=1.60$ N. Unlike charges, so it pulls $q_0$ towards $B$: $\mathbf F_2=1.60\,(0.8,\,-0.6)=(1.28,\ -0.96)$ N.
Add $F_x=1.28$ N, $F_y=3.33-0.96=2.37$ N. $$|\mathbf F|=\sqrt{1.28^2+2.37^2}\approx2.70\ \text{N},\qquad\theta=\tan^{-1}\frac{2.37}{1.28}\approx61.7^\circ$$ above the $+x$ direction (the direction from $A$ to $B$).
Check Interactive 2 opens on exactly this arrangement: watch the two forces slide tip-to-tail and the readouts give $F_x=1.28$, $F_y=2.37$, $|\mathbf F|=2.70$ N. ✓
✏️ Example 6 — Where is the net force zero?
Charges $+4.0\ \mu\text{C}$ and $+1.0\ \mu\text{C}$ are fixed $12$ cm apart. (a) Where on the line joining them does a third charge feel no net force? Does the answer depend on the third charge? (b) Repeat if the second charge is $-1.0\ \mu\text{C}$.
(a) Where? Put $+4\ \mu\text{C}$ at $x=0$ and $+1\ \mu\text{C}$ at $x=12$ cm. For like charges the two forces on a third charge point in opposite directions only between them, so the point lies there, at some $x$ with $0\lt x\lt12$ cm.
(a) Solve Equal magnitudes: $\dfrac{k(4\ \mu\text{C})q_0}{x^2}=\dfrac{k(1\ \mu\text{C})q_0}{(12-x)^2}$, so $\dfrac{12-x}{x}=\dfrac12$ (taking positive roots), giving $x=8.0$ cm from the $+4\ \mu\text{C}$ charge. $q_0$ cancels: any third charge, of either sign, feels zero force there.
(b) Unlike charges Between unlike charges both forces point the same way, so the zero must be outside, on the side of the smaller charge (so that the larger charge is farther away). With $x\gt12$ cm: $\dfrac{4}{x^2}=\dfrac{1}{(x-12)^2}\Rightarrow x-12=\tfrac{x}{2}\Rightarrow x=24$ cm, i.e. $12$ cm beyond the $-1\ \mu\text{C}$ charge.
Check In (a): $F$ from $4\ \mu\text{C}$ at $8$ cm $\propto 4/64=1/16$; from $1\ \mu\text{C}$ at $4$ cm $\propto1/16$. ✓ Choose “neutral point” in Interactive 2 and drag $q_0$ along the line to see the net force vanish at $8$ cm.
⚠️ Common mistakes with Coulomb's law and superposition

1. Forgetting to square $r$, or to convert units. Centimetres and microcoulombs must become metres and coulombs before you use $k=9.0\times10^9$.

2. Adding magnitudes instead of vectors. In Example 5, $3.33+1.60=4.93$ N is wrong; the forces point in different directions and the correct net force is $2.70$ N.

3. Using signs in the magnitude formula and then again for direction. Use $|q_1q_2|$ for the size and decide the direction from a diagram — or use the vector form consistently, but never mix the two.

4. Applying $kq_1q_2/r^2$ to large bodies. The law is for point charges. For extended charged bodies, superposition over small pieces is needed (Section 1.3); uniformly charged spheres are a special case that act like point charges at their centres, from outside.

Interactive 2 — Superposition Lab: Adding Coulomb Forces

Each coloured arrow is the force on the test charge $q_0$ from one source charge alone. Press Play to watch them slide tip-to-tail; the thick arrow from $q_0$ to the final tip is the net force $\mathbf F=\sum\mathbf F_i$. Drag any charge (grid squares are $1$ cm) and watch every arrow update.

+1
individual forces—
$F_x$ (N)—
$F_y$ (N)—
$|\mathbf F|$ (N)—
direction $\theta$ from $+x$—

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.

Exercises

I
Charge and its properties. Take $e=1.6\times10^{-19}$ C and $m_e=9.1\times10^{-31}$ kg.
  • 1 How many electrons must be removed from a neutral coin to give it a charge of $+1.0\ \mu\text{C}$?
  • 2 Which of these charges can a body carry: $4.8\times10^{-19}$ C, $2.0\times10^{-19}$ C, $-8.0\times10^{-19}$ C, $1.6\times10^{-18}$ C? Give reasons.
  • 3 A plastic comb pulled through dry hair gains $2.0\times10^{9}$ electrons. What are the charges on the comb and on the hair?
  • 4 Two identical metal spheres carry $+7.0\ \mu\text{C}$ and $-3.0\ \mu\text{C}$. They touch and are separated. What is the charge on each?
  • 5 Explain why a rubbed comb attracts small uncharged pieces of paper, and why some pieces then jump away after touching it.
  • 6 A negatively charged rod is used to charge an insulated metal sphere by induction. List the steps, and state the sign of the final charge on the sphere.
  • 7 One mole of water (18 g) contains $6.02\times10^{23}$ molecules, each with 10 protons. Find the total positive charge in it. Why do we never notice this charge?
  • 8 By how much does the mass of a body change when it is given a charge of $-1.0\ \mu\text{C}$?
1 $n=\dfrac{q}{e}=\dfrac{1.0\times10^{-6}}{1.6\times10^{-19}}=6.25\times10^{12}\approx6.3\times10^{12}$ electrons.
2 Divide by $e$: $4.8\times10^{-19}=3e$ ✓; $2.0\times10^{-19}=1.25e$ ✗ (not an integer); $-8.0\times10^{-19}=-5e$ ✓; $1.6\times10^{-18}=10e$ ✓.
3 $q=-ne=-(2.0\times10^9)(1.6\times10^{-19})=-3.2\times10^{-10}$ C on the comb; the hair has $+3.2\times10^{-10}$ C (conservation).
4 Total $+4.0\ \mu\text{C}$ shared equally: $+2.0\ \mu\text{C}$ on each.
5 The comb polarises the paper: charges in its molecules shift slightly, so the side nearer the comb carries opposite charge. The nearer charge is attracted more strongly than the farther charge is repelled, giving a net attraction. A piece that touches the comb picks up some of the comb's charge, now has the same sign, and is repelled.
6 (i) Bring the rod near the sphere: electrons are pushed to the far side. (ii) Earth the sphere: these electrons flow to the Earth. (iii) Remove the earth connection. (iv) Remove the rod. The sphere is left positive.
7 $Q=10\times(6.02\times10^{23})\times(1.6\times10^{-19})\approx9.6\times10^{5}$ C — almost a million coulombs. The water also contains exactly as many electrons, so the net charge is zero and no force is felt.
8 It gains $6.25\times10^{12}$ electrons: $\Delta m=(6.25\times10^{12})(9.1\times10^{-31})\approx5.7\times10^{-18}$ kg (an increase).
II
Coulomb's law for two charges. Use $k=9.0\times10^{9}\ \text{N m}^2\,\text{C}^{-2}$ and give directions.
  • 1 Find the force between $+1.0\ \mu\text{C}$ and $+4.0\ \mu\text{C}$ placed $20$ cm apart in vacuum.
  • 2 How far apart must two electrons be for the repulsion between them to equal the weight of one electron? ($g=9.8\ \text{m s}^{-2}$)
  • 3 Two charges attract with force $F$. Each charge is doubled and the separation is tripled. What is the new force?
  • 4 Two equal point charges $5.0$ cm apart in air repel with $0.90$ N. Find each charge.
  • 5 Two charges at a fixed separation repel with $12$ N in air and $4.8$ N when immersed in an oil. Find the relative permittivity of the oil.
  • 6 Use Coulomb's law to find the SI unit and the dimensional formula of $\varepsilon_0$.
  • 7 A charge $Q$ is divided into two parts $q$ and $Q-q$, which are then placed a fixed distance apart. For what value of $q$ is the repulsion greatest? (uses differentiation)
  • 8 Nuclear physics. In a scattering experiment an alpha particle (charge $+2e$, mass $6.64\times10^{-27}$ kg) comes within $5.0\times10^{-14}$ m of a gold nucleus (charge $+79e$). Find the force on the alpha particle and its acceleration at that moment.
Check: a Coulomb's-law calculator for your answers
$|F|$ (N)—
nature—
$q_1/e$ and $q_2/e$—
1 $F=\dfrac{(9.0\times10^9)(1.0\times10^{-6})(4.0\times10^{-6})}{(0.20)^2}=0.90$ N, repulsive.
2 $\dfrac{ke^2}{r^2}=m_eg\Rightarrow r=e\sqrt{\dfrac{k}{m_eg}}=(1.6\times10^{-19})\sqrt{\dfrac{9.0\times10^9}{(9.1\times10^{-31})(9.8)}}\approx5.1$ m. Even this tiny charge beats gravity at metre distances.
3 $F'=\dfrac{k(2q_1)(2q_2)}{(3r)^2}=\dfrac49F$, still attractive.
4 $q=r\sqrt{\dfrac{F}{k}}=(0.050)\sqrt{\dfrac{0.90}{9.0\times10^9}}=(0.050)(1.0\times10^{-5})=5.0\times10^{-7}$ C $=0.50\ \mu\text{C}$ each (both of the same sign).
5 $\varepsilon_r=\dfrac{F_\text{air}}{F_\text{oil}}=\dfrac{12}{4.8}=2.5$.
6 $\varepsilon_0=\dfrac{q_1q_2}{4\pi Fr^2}$, so its unit is $\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$. With $[q]=[\text{AT}]$ and $[F]=[\text{MLT}^{-2}]$: $[\varepsilon_0]=\dfrac{[\text{A}^2\text{T}^2]}{[\text{MLT}^{-2}][\text{L}^2]}=[\text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{A}^{2}]$.
7 $F(q)=\dfrac{k\,q(Q-q)}{r^2}$. $\dfrac{dF}{dq}=\dfrac{k(Q-2q)}{r^2}=0\Rightarrow q=\dfrac Q2$, and $\dfrac{d^2F}{dq^2}=-\dfrac{2k}{r^2}\lt0$, so this is a maximum: split the charge equally.
8 $F=\dfrac{k(2e)(79e)}{r^2}=\dfrac{(9.0\times10^9)(158)(1.6\times10^{-19})^2}{(5.0\times10^{-14})^2}\approx14.6$ N, repulsive (away from the nucleus). $a=\dfrac Fm=\dfrac{14.6}{6.64\times10^{-27}}\approx2.2\times10^{27}\ \text{m s}^{-2}$. A force you could feel with your hand, acting on a particle of almost no mass — it is turned back in a tiny fraction of a second (Chapter 12).
III
Superposition. Distances are in centimetres; give magnitudes and directions.
  • 1 Charges $+2.0\ \mu\text{C}$ at $x=0$ and $-2.0\ \mu\text{C}$ at $x=10$ cm. Find the force on $+1.0\ \mu\text{C}$ at $x=5$ cm.
  • 2 Same two charges; find the force on $+1.0\ \mu\text{C}$ at $x=15$ cm.
  • 3 Three charges of $+2.0\ \mu\text{C}$ sit at the corners of an equilateral triangle of side $10$ cm. Find the net force on any one of them.
  • 4 Four charges of $+1.0\ \mu\text{C}$ sit at the corners of a square of side $10$ cm. Find the net force on any one of them.
  • 5 Charges $+9.0\ \mu\text{C}$ at $x=0$ and $+1.0\ \mu\text{C}$ at $x=20$ cm. Where on the $x$-axis is the net force on a third charge zero?
  • 6 Repeat part 5 with the second charge equal to $-1.0\ \mu\text{C}$.
  • 7 Charges of $+3.0\ \mu\text{C}$ are fixed at $(0,\,3\ \text{cm})$ and $(0,\,-3\ \text{cm})$. Find the force on $+1.0\ \mu\text{C}$ at $(4\ \text{cm},\,0)$.
  • 8 Challenge. In part 7, let the $+1.0\ \mu\text{C}$ charge move along the positive $x$-axis. Show that the force is $F(x)=\dfrac{2kqq_0\,x}{(x^2+a^2)^{3/2}}$ with $a=3$ cm, and find where it is greatest and its greatest value.
1 Each charge is $5$ cm away: $F=\dfrac{(9.0\times10^9)(2.0\times10^{-6})(1.0\times10^{-6})}{(0.05)^2}=7.2$ N. The $+2$ pushes it to $+x$, the $-2$ pulls it to $+x$: net $14.4$ N along $+x$.
2 From $+2\ \mu\text{C}$ ($15$ cm): $0.80$ N along $+x$. From $-2\ \mu\text{C}$ ($5$ cm): $7.2$ N along $-x$. Net $6.4$ N along $-x$.
3 Each pair: $F=\dfrac{(9.0\times10^9)(2.0\times10^{-6})^2}{(0.10)^2}=3.6$ N, at $60^\circ$ to each other. Net $=2F\cos30^\circ=\sqrt3(3.6)\approx6.2$ N, directed away from the centre of the triangle along the bisector of the corner. (Interactive 2, arrangement 2.)
4 Adjacent charges ($10$ cm): $0.90$ N each, at right angles, resultant $0.90\sqrt2=1.27$ N along the diagonal. Opposite corner ($10\sqrt2$ cm): $\dfrac{(9.0\times10^9)(10^{-12})}{0.02}=0.45$ N along the same diagonal. Net $1.27+0.45\approx1.72$ N, outwards along the diagonal. (Arrangement 3.)
5 Between the charges: $\dfrac{9}{x^2}=\dfrac{1}{(20-x)^2}\Rightarrow3(20-x)=x\Rightarrow x=15$ cm.
6 Outside, beyond the smaller charge: $\dfrac{9}{x^2}=\dfrac{1}{(x-20)^2}\Rightarrow3(x-20)=x\Rightarrow x=30$ cm.
7 Each source is $\sqrt{4^2+3^2}=5$ cm away: $F=\dfrac{(9.0\times10^9)(3.0\times10^{-6})(1.0\times10^{-6})}{(0.05)^2}=10.8$ N. The $y$-components cancel by symmetry; each $x$-component is $10.8\times\tfrac45=8.64$ N. Net $17.3$ N along $+x$.
8 Each source is $\sqrt{x^2+a^2}$ away and contributes $x$-component $\dfrac{kqq_0}{x^2+a^2}\cdot\dfrac{x}{\sqrt{x^2+a^2}}$; doubling gives $F(x)$. Setting $F'(x)=0$: $(x^2+a^2)-3x^2=0\Rightarrow x=\dfrac{a}{\sqrt2}\approx2.1$ cm. There $F_\text{max}=\dfrac{2kqq_0}{a^2}\cdot\dfrac{2}{3\sqrt3}=2(30\ \text{N})(0.385)\approx23.1$ N, using $\dfrac{kqq_0}{a^2}=\dfrac{(9.0\times10^9)(3.0\times10^{-12})}{(0.03)^2}=30$ N.
IV
Thinking with Coulomb's law.
  • 1 Modelling. Two small balls, each of mass $1.0$ g, hang from the same point on insulating threads $50$ cm long. Given equal charges, they come to rest $6.0$ cm apart. Find the charge on each ball. ($g=9.8\ \text{m s}^{-2}$)
  • 2 Two $1.0$ kg masses are given equal charges $q$ so that their electric repulsion exactly balances their gravitational attraction. Find $q$ and the number of extra electrons this represents. ($G=6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$)
  • 3 The electric force is about $10^{39}$ times stronger than gravity between a proton and an electron, yet in everyday life we feel gravity but rarely notice electric forces. Explain.
  • 4 Charges $+q$ and $+4q$ are held a distance $L$ apart. Find the position and value of a third charge $Q$ that puts all three charges in equilibrium.
1 Each thread makes angle $\theta$ with the vertical, $\sin\theta=\tfrac{3.0}{50}$, so $\tan\theta=\dfrac{3.0}{\sqrt{50^2-3.0^2}}\approx0.0601$. Equilibrium: $F=mg\tan\theta=(1.0\times10^{-3})(9.8)(0.0601)\approx5.9\times10^{-4}$ N. Then $q=r\sqrt{F/k}=(0.060)\sqrt{\dfrac{5.9\times10^{-4}}{9.0\times10^{9}}}\approx1.5\times10^{-8}$ C $\approx15$ nC.
2 $\dfrac{kq^2}{r^2}=\dfrac{Gm^2}{r^2}\Rightarrow q=m\sqrt{\dfrac Gk}=(1.0)\sqrt{\dfrac{6.67\times10^{-11}}{9.0\times10^{9}}}\approx8.6\times10^{-11}$ C, i.e. about $5.4\times10^{8}$ electrons — out of roughly $10^{26}$ electrons in a kilogram of matter. The separation does not matter.
3 Matter is almost exactly neutral: every body contains (very nearly) equal amounts of positive and negative charge, so the enormous electric forces cancel. Gravity has only one sign (always attractive), so it adds up over the $10^{50}$ particles of the Earth. The electric forces we do meet — friction, contact forces, tension — are the leftover, short-range effects of charges in neighbouring atoms.
4 For $Q$ to be in equilibrium it must lie between the charges at distance $x$ from $+q$: $\dfrac{q}{x^2}=\dfrac{4q}{(L-x)^2}\Rightarrow L-x=2x\Rightarrow x=\dfrac L3$. For $+q$ to be in equilibrium, the pull of $Q$ must balance the push of $+4q$: $\dfrac{k|Q|q}{(L/3)^2}=\dfrac{k(4q)q}{L^2}\Rightarrow|Q|=\dfrac{4q}{9}$, and $Q$ must attract, so $Q=-\dfrac{4q}{9}$ at $L/3$ from $+q$. (Then $+4q$ is in equilibrium automatically, since the net force on the whole system is zero.)
🎯 Exam corner — charge and Coulomb's law BoardJEE MainJEE AdvancedNEET

E1. Board (a) Assertion (A): No object can carry a charge of $3.2\times10^{-20}$ C. Reason (R): The charge on any body is an integral multiple of $1.6\times10^{-19}$ C. 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. (b) A glass rod rubbed with silk acquires $+4.8\times10^{-8}$ C. How many electrons were transferred, and from which body to which? (1 + 2 marks)

Concept Quantisation: $q=ne$ with $n$ an integer; rubbing transfers electrons only.
(a) $\dfrac{3.2\times10^{-20}}{1.6\times10^{-19}}=0.2$, not an integer, so A is true; R is true and is exactly the reason. Answer (A).
(b) $n=\dfrac{4.8\times10^{-8}}{1.6\times10^{-19}}=3.0\times10^{11}$ electrons, transferred from the glass rod to the silk (the rod is left positive; the silk gets $-4.8\times10^{-8}$ C).
Exam tip Board examiners give the mark for the direction of transfer separately — always say which body lost electrons.

E2. Board State Coulomb's law in vector form and use it to show that the forces two point charges exert on each other obey Newton's third law. A charge $q_1=+5.0$ nC is at the origin and $q_2=-2.0$ nC is at $(3.0\ \text{cm},\ 4.0\ \text{cm})$. Find the force on $q_2$ in component form. (3 marks)

Law $\mathbf F_{21}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}\hat{\mathbf r}_{21}$, where $\hat{\mathbf r}_{21}$ points from $q_1$ to $q_2$. For the force on $q_1$, swap labels: $\mathbf F_{12}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}\hat{\mathbf r}_{12}$, and $\hat{\mathbf r}_{12}=-\hat{\mathbf r}_{21}$, so $\mathbf F_{12}=-\mathbf F_{21}$: equal and opposite, along the same line. ✓
Setup $\mathbf r_{21}=(0.030,\,0.040)$ m, $r=0.050$ m, $\hat{\mathbf r}_{21}=(0.6,\,0.8)$.
Equation $\dfrac{kq_1q_2}{r^2}=\dfrac{(9.0\times10^9)(5.0\times10^{-9})(-2.0\times10^{-9})}{(0.050)^2}=-3.6\times10^{-5}$ N, so $$\mathbf F_{21}=-3.6\times10^{-5}\,(0.6,\,0.8)\ \text{N}=\left(-2.16\times10^{-5},\ -2.88\times10^{-5}\right)\ \text{N}.$$
Answer Magnitude $3.6\times10^{-5}$ N, directed from $q_2$ towards the origin (attraction). Exam tip: the negative sign in the vector form does the direction work for you — do not add a second minus sign by hand.

E3. JEE Main Two identical small conducting spheres carry charges $+12\ \mu\text{C}$ and $-4\ \mu\text{C}$ and attract each other with a force $F$ when a distance $r$ apart. They are touched together and then placed a distance $r/2$ apart. The new force is $\dfrac{x}{3}F$. Find the integer $x$.

Concept Identical conductors share the total charge equally; then Coulomb's law.
Before $F=\dfrac{k(12)(4)}{r^2}=\dfrac{48k}{r^2}$ (in $\mu\text{C}^2$ units), attractive.
After Each sphere has $\tfrac{12-4}{2}=+4\ \mu\text{C}$, so $F'=\dfrac{k(4)(4)}{(r/2)^2}=\dfrac{64k}{r^2}$, now repulsive.
Answer $\dfrac{F'}{F}=\dfrac{64}{48}=\dfrac43$, so $x=\mathbf 4$. Exam tip: JEE likes to change both the charges and the distance; write the ratio $F'/F$ directly and let $k$ cancel.

E4. NEET Two point charges in air repel with force $F$ when a distance $d$ apart. They are immersed in a liquid of dielectric constant $4$ and their separation is changed to $d/2$. The force is now (A) $F/4$ (B) $F$ (C) $4F$ (D) $16F$.

Concept $F\propto\dfrac{1}{\varepsilon_r r^2}$.
Ratio $\dfrac{F'}{F}=\dfrac{1}{\varepsilon_r}\left(\dfrac{d}{d/2}\right)^2=\dfrac14\times4=1$.
Answer (B) $F$. Exam tip: the two effects cancel exactly — a favourite NEET trap is to apply only one of them and pick (A) or (C).

E5. JEE Advanced (One or more correct.) Two identical small balls of density $1600\ \text{kg m}^{-3}$ carry equal charges and hang from a common point on light insulating threads of equal length; in air each thread makes an angle $\theta$ with the vertical. The whole arrangement is then immersed in a non-conducting liquid of density $800\ \text{kg m}^{-3}$, and the angle is found to be unchanged. Which statements are correct? (A) The dielectric constant of the liquid is $2$. (B) The tension in each thread in the liquid is half its value in air. (C) The electric force between the balls in the liquid is half its value in air. (D) If the liquid had the same density but dielectric constant $1$, the angle would decrease.

Setup Each ball is in equilibrium under its effective weight $W$ (downwards), the thread tension $T$ and the horizontal electric force $F$: $T\cos\theta=W$, $T\sin\theta=F$, so $\tan\theta=F/W$.
Buoyancy In the liquid $W'=mg\left(1-\dfrac{\rho_\ell}{\rho_b}\right)=mg\left(1-\dfrac{800}{1600}\right)=\dfrac{mg}{2}$.
Options Same $\theta$ ⇒ $F'/W'=F/W$ ⇒ $F'=F/2$, so (C) is true, and since the separation is unchanged $F'=F/\varepsilon_r$ gives $\varepsilon_r=2$: (A) true. $T'=W'/\cos\theta=T/2$: (B) true. With $\varepsilon_r=1$, $F$ would stay the same while $W$ halves, so $\tan\theta$ would increase and the threads would spread: (D) false.
Answer (A), (B), (C). Exam tip: in a liquid always correct the weight for buoyancy and the Coulomb force for $\varepsilon_r$; then compare equilibrium equations rather than computing numbers.

E6. JEE Main NEET Three charges, each $+3.0\ \mu\text{C}$, sit at the corners of an equilateral triangle. What charge $Q$ placed at the centroid makes the net force on every charge zero? (A) $+1.7\ \mu\text{C}$ (B) $-1.7\ \mu\text{C}$ (C) $-3.0\ \mu\text{C}$ (D) $-5.2\ \mu\text{C}$

Concept Superposition at one corner; by symmetry the other corners follow.
Corner forces Side $a$: the two repulsions $kq^2/a^2$ are $60^\circ$ apart, so their resultant is $2\dfrac{kq^2}{a^2}\cos30^\circ=\sqrt3\,\dfrac{kq^2}{a^2}$, directed away from the centroid.
Balance The centroid is $a/\sqrt3$ from each corner, so $Q$ exerts $\dfrac{k|Q|q}{a^2/3}=\dfrac{3k|Q|q}{a^2}$, and it must pull towards the centroid, so $Q$ is negative: $3|Q|=\sqrt3\,q\Rightarrow|Q|=\dfrac{q}{\sqrt3}=\dfrac{3.0}{1.732}\approx1.7\ \mu\text{C}$.
Answer (B) $Q=-q/\sqrt3\approx-1.7\ \mu\text{C}$, independent of the side length. Exam tip: the net force on $Q$ itself is zero by symmetry, but this equilibrium is unstable — a slight push and it collapses.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How can we tell, from measurements alone, that electric charge comes in whole-number packets — and how precisely can a small data set pin down the size of the packet $e$?

Why it matters. Between 1909 and 1913 Robert Millikan and Harvey Fletcher in Chicago balanced tiny charged oil drops between charged plates and found that every drop's charge was a multiple of one smallest value. Today the same idea is used in reverse: "single-electron" devices built in laboratories in Europe, Japan and the USA move electrons one at a time, and since 2019 the SI defines the coulomb by fixing $e$ exactly at $1.602176634\times10^{-19}$ C. Counting charge is now how current is measured at the highest precision.

What to do.

  1. Sample data — charges measured on eight drops, in units of $10^{-19}$ C: $6.41,\ 9.62,\ 4.79,\ 12.83,\ 8.01,\ 3.22,\ 16.04,\ 11.20$.
  2. Sort them and look at the differences between neighbouring values. What is the smallest difference that keeps reappearing?
  3. Guess a value of $e$, divide every charge by it and round to whole numbers $n$. Plot $q$ against $n$; the points should lie on a straight line through the origin.
  4. Estimate $e$ as the slope of that line (or as $\sum q/\sum n$) and compare with the defined value. Then explain why the data can never prove that $e$ is not half as large — and what extra evidence would settle it.

Tools: Calculator or spreadsheet; graph paper.

Hand in: Your table of $q$ and $n$, the graph, your estimate of $e$ with its percentage difference from $1.602\times10^{-19}$ C, and a short paragraph on the "half-as-large" question.

Going further. Read about how a single-electron pump produces a current $I=ef$ by moving one electron per cycle at frequency $f$. What current does a $1$ GHz pump deliver, and why is that useful to a national standards laboratory?

Instructor note: the data correspond to $n=4,6,3,8,5,2,10,7$; the estimate is $\sum q/\sum n=72.12/45\approx1.603\times10^{-19}$ C (about $0.05\%$ high). Smaller packets ($e/2$, $e/3$) would fit equally well, which is exactly why many drops with small $n$ are needed — the smallest observed differences rule out larger units but never smaller ones. A $1$ GHz pump gives $I=(1.6\times10^{-19})(10^9)\approx0.16$ nA.

🧪 Self-check quiz

0/5 correct

Q1. Which of these charges cannot be found on an isolated body?

Q2. The distance between two point charges is halved. The force between them becomes

Q3. A positively charged rod is brought near (not touching) an uncharged, insulated metal sphere. The sphere is

Q4. Identical metal spheres with charges $+5\ \mu\text{C}$ and $-1\ \mu\text{C}$ are touched and separated. Each now carries

Q5. Charges $+q$ at $x=0$ and $+4q$ at $x=3a$. A third charge feels zero net force at

← Chapter 1 — Electric Charges and Fields1.2 The Electric Field, Field Lines and Electric Dipoles →