Welcome to Class 12 Physics. Take a breath — you are starting the very first chapter of the course, and it is a kind one. Electric Charges and Fields asks you to accept only a few fundamental behaviours of charge at the start — that it exists in two flavours, and that it is conserved and quantised — and that charges push and pull each other with a force you can actually calculate. Everything else in this chapter grows out of those two ideas, step by step, like a staircase. If Class 11 mechanics ever felt like a wall of formulas, you will find this chapter friendlier, because almost every result comes from one single equation — Coulomb’s law — dressed up in different clothes.
In the CBSE board paper, Electrostatics (this chapter plus Electrostatic Potential and Capacitance) sits in a block worth 16 marks that Electrostatics shares with Current Electricity — those 16 marks are spread across the first three chapters, not this unit alone — and this chapter by itself typically contributes around 6–8 marks. Almost every year there is a derivation from Gauss’s theorem — the field of a long wire, an infinite sheet, or a spherical shell — and those are the cheapest full marks in the whole paper, because the examiner is looking for a fixed sequence of steps. Learn that sequence and you have banked 5 marks before you even open the numerical section.
We are going to build this from absolute zero. No prior electrostatics assumed. Read slowly, do the worked examples with a pen in your hand, and do not move on from a section until it feels comfortable. That last part matters more than speed.
- What Electric Charge Really Is
- The Three Basic Properties of Charge
- How Objects Get Charged: Friction, Conduction and Induction
- Coulomb’s Law and Its Vector Form
- Permittivity and the Effect of a Medium
- Forces Between Multiple Charges: The Superposition Principle
- Electric Field and the Field of a Point Charge
- Electric Field Lines and Their Properties
- Electric Dipole and Dipole Moment
- Field of a Dipole: Axial and Equatorial Lines
- Torque on a Dipole in a Uniform Field
- Continuous Charge Distributions: λ, σ and ρ
- Electric Flux
- Gauss’s Theorem
- Application 1: Infinitely Long Straight Charged Wire
- Application 2: Uniformly Charged Infinite Plane Sheet
- Application 3: Uniformly Charged Thin Spherical Shell
- Formula Sheet at a Glance
- Practice Worksheet
Your Game Plan
Chapters go wrong when students jump straight to numericals. Here is the order that actually works for this one:
- Day 1 — the language. Charge, its properties, and Coulomb’s law. Do nothing but two-charge problems until they feel automatic.
- Day 2 — superposition. Three and four charges. This is pure Class 11 vector addition wearing an electrostatics costume. Draw every force before you compute anything.
- Day 3 — the field idea. Move from “force between charges” to “field created by a charge”. Learn field lines properly; one-mark questions live here.
- Day 4 — the dipole. Dipole moment, axial field, equatorial field, torque. Memorise the factor-of-2 relationship and you are safe.
- Day 5 — flux and Gauss. Then the three standard derivations, written out from memory, three times each. Not read — written.
- Day 6 — the worksheet at the bottom of this page, closed book, timed.
Start a single sheet titled “Gauss derivations”. Every time you write out the wire, sheet or shell derivation, date it. By the third repetition you will be writing it in under four minutes, which is exactly what the board wants for a 5-mark question.
What Electric Charge Really Is
You already know charge from everyday life, even if nobody called it that. Comb your dry hair and the comb picks up tiny bits of paper. Pull off a sweater in winter and you hear faint crackles. Touch a car door after a long drive and you get a small shock. In every one of those cases, something invisible moved from one object to another and left both objects in an unusual state. That “something” is electric charge.
Charge is not a substance you can put in a jar. It is a property that some particles carry, in the same way that mass is a property. The difference is that mass comes in only one kind — there is no negative mass — while charge comes in two kinds, which we call positive and negative. The labels are just historical bookkeeping; nothing is intrinsically “more” or “less” about either.
The rule that follows from having two kinds is the one you must never forget: like charges repel, unlike charges attract. Two positives push apart. Two negatives push apart. A positive and a negative pull together.
Inside an atom, the protons in the nucleus carry positive charge and the electrons around it carry an equal negative charge. A normal atom has equal numbers of each, so its total charge is zero — we say it is neutral. Neutral does not mean “no charge inside”; it means “the positive and negative exactly cancel”. That distinction will matter a lot when we get to induction.
The SI unit of charge is the coulomb (C). It is defined through current: one coulomb is the charge that flows past a point in one second when the current is one ampere, so 1 C = 1 A s. One coulomb is an enormous amount of charge for static electricity. The charge on that comb in your hand is more like a microcoulomb, a millionth of a coulomb. This is why almost every number in this chapter is written with μC (10−6 C) or nC (10−9 C).
Electron: q = −e = −1.6 × 10−19 C.
Proton: q = +e = +1.6 × 10−19 C.
The symbol e always stands for the positive number 1.6 × 10−19 C. The minus sign for the electron is written separately.
How big is one coulomb, really? We will answer that properly in the next section, but here is a preview: it takes about 6.25 × 1018 electrons to make one coulomb. If you could count one electron per second, you would need roughly two hundred billion years. That is the scale we are dealing with.
The Three Basic Properties of Charge
Charge obeys three rules that the board asks about again and again. They are short, so learn them word-perfect.
1. Additivity. Charge is a scalar, and the total charge on a body is the plain algebraic sum of all the individual charges on it, signs included. If a body carries +5 μC, −3 μC and +2 μC at three different points, the total is (+5) + (−3) + (+2) = +4 μC. Notice that you add them like ordinary numbers — no angles, no components. Forces between charges are vectors; charge itself is not.
2. Quantisation. Charge does not come in a continuous stream. It comes in whole packets of size e = 1.6 × 10−19 C. So any observable charge must be
q = n e, where n = 0, ±1, ±2, ±3, …
Think of money. You cannot pay 3.7 paise; the smallest coin fixes the step size. Here e is the smallest coin. A charge of 2.4 × 10−19 C simply cannot exist on a body, because 2.4 ÷ 1.6 = 1.5, and you cannot have one and a half electrons.
3. Conservation. Charge can be moved around, but it cannot be created or destroyed. In any isolated system, the total charge before an event equals the total charge after it. When you rub a glass rod with silk, the rod becomes positive and the silk becomes equally negative — the pair still adds to zero. Charge was transferred, not manufactured.
Additive: total charge is the algebraic sum, because charge is a scalar.
Quantised: q = ne, so charge always comes in whole multiples of 1.6 × 10−19 C.
Conserved: total charge of an isolated system never changes.
Step 1 — convert to SI. q = −1 μC = −1 × 10−6 C.
Step 2 — use quantisation. The charge is negative, so the bead has extra electrons and q = −ne. Taking magnitudes,
n = |q| ÷ e = (1 × 10−6) ÷ (1.6 × 10−19)
Step 3 — do the arithmetic carefully. 1 ÷ 1.6 = 0.625, and 10−6 ÷ 10−19 = 1013.
n = 0.625 × 1013 = 6.25 × 1012 electrons.
Sense check: that is about six trillion electrons for a charge you would call “tiny”. Now you can see why one whole coulomb is so large: it needs 6.25 × 1018 electrons, a million times more again.
Method: divide by e and see whether you get a whole number.
(a) n = (3.2 × 10−20) ÷ (1.6 × 10−19) = 2 × 10−1 = 0.2
0.2 is not an integer, so this charge is not possible. It would need one-fifth of an electron.
(b) n = (4.8 × 10−19) ÷ (1.6 × 10−19) = 3
A whole number, so this charge is possible — the body has lost 3 electrons (if positive) or gained 3 electrons (if negative).
Why it works: quantisation says every real charge is an exact multiple of e. Dividing by e is simply asking “how many coins is this?” If the answer has a decimal, the amount cannot be paid.
Why quantisation almost never shows up in daily life. Because e is so ridiculously small, a charge of 1 μC is already 6.25 × 1012 steps. Adding or removing one electron changes it by one part in six trillion — far below anything you could measure. So at the macroscopic scale charge looks perfectly continuous, exactly the way a beach looks like smooth sand until you get a microscope out and find individual grains. Quantisation only becomes visible at the atomic scale.
Students who have just finished Class 11 mechanics sometimes try to add charges using components or the parallelogram law. Do not. Charge is a scalar. You add +5 μC and −3 μC to get +2 μC, full stop. It is the force between charges that is a vector.
How Objects Get Charged: Friction, Conduction and Induction
Under the rationalised NCERT text, the detailed treatment of conductors and insulators and of charging by induction was trimmed from the Class 12 chapter, and the CBSE course statement for Unit I now begins at “Electric charges, conservation of charge, Coulomb’s law”. We have kept this section because the idea explains a lot of later physics and appears in competitive exams, but do confirm against your own school’s current syllabus copy before spending heavy revision time on it.
Before any of the maths, it helps to know how a body ends up charged in the first place. There are three routes.
Charging by friction (rubbing). When two different materials are rubbed together, electrons are transferred from the one that holds them loosely to the one that holds them tightly. Rub a glass rod with silk: electrons leave the glass and settle on the silk, so glass goes positive and silk goes equally negative. Rub an ebonite rod with fur and it goes the other way — ebonite goes negative. Notice the pattern: only electrons move. Protons are locked inside nuclei and never travel in these experiments. A body becomes positive by losing electrons, not by gaining protons.
Charging by conduction (contact). Touch a charged conductor to an uncharged one and charge simply spreads across both. If the two bodies are identical conducting spheres, the charge shares out equally, because there is nothing to make one sphere preferred over the other. This gives a very common exam step: two identical spheres with charges q1 and q2 touch and are separated, and each ends up with (q1 + q2) ÷ 2. Note that the algebraic sum is used, so signs must be carried through.
Charging by induction (no contact at all). This is the clever one. Bring a positively charged rod near a neutral metal sphere on an insulating stand, without touching it. The free electrons that make metals good conductors are attracted towards the rod and crowd onto the near face, leaving the far face short of electrons and therefore positive. The sphere is still neutral overall — charge has only been rearranged. Now earth the far face for a moment: electrons flow up from the earth to neutralise that positive far side. Remove the earth connection first, then take the rod away. The sphere is left with a net negative charge.
Two things are worth noticing. First, the induced charge is opposite in sign to the inducing charge. Second, the rod never lost any charge of its own — you got a charged sphere for free, which is why induction can be repeated endlessly with the same rod. The order of operations is what students get wrong: earth first, remove the earth, then remove the rod. Reverse the last two and the sphere goes back to neutral.
| Basis | Friction | Conduction | Induction |
|---|---|---|---|
| Contact needed? | Yes, rubbing | Yes, touching | No |
| Sign acquired by the target body | Opposite to the rubbing partner | Same as the source | Opposite to the source |
| Does the source lose charge? | Yes (both bodies change) | Yes, it shares | No |
| Works on insulators? | Yes | Poorly | Only weakly (polarisation) |
Step 1 — apply conservation and additivity. Total charge before contact = (+8) + (−2) = +6 μC. Conservation says the total is still +6 μC after contact.
Step 2 — share it out. The spheres are identical, so each takes half:
q = (+6 μC) ÷ 2 = +3 μC on each sphere.
Step 3 — now use Coulomb’s law (introduced fully in the next section):
F = k q1q2 ÷ r² = (8.99 × 109)(3 × 10−6)(3 × 10−6) ÷ (0.20)²
Numerator: (8.99 × 109)(9 × 10−12) = 8.091 × 10−2
Denominator: (0.20)² = 0.040
F = (8.091 × 10−2) ÷ 0.040 = 2.02 N, repulsive.
Note the sign story: the spheres started out attracting each other and ended up repelling. Both are now positive.
With +8 μC and −2 μC, students often write (8 + 2) ÷ 2 = 5 μC. Wrong. Use the algebraic sum: (+8 − 2) ÷ 2 = +3 μC. Carry the minus sign all the way into the division.
Coulomb’s Law and Its Vector Form
Here is the engine of the whole chapter. Charles Coulomb measured the force between two small charged spheres and found something that should feel familiar from gravitation: the force falls off as the square of the distance.
Statement. The force between two stationary point charges is directly proportional to the product of their magnitudes, inversely proportional to the square of the distance between them, and acts along the straight line joining them.
F = (1 ÷ 4πε0) × q1q2 ÷ r²
The constant is usually written k = 1 ÷ (4πε0) = 8.99 × 109 N m² C−2, often rounded to 9 × 109 in exams. The quantity ε0 is the permittivity of free space, ε0 = 8.854 × 10−12 C² N−1 m−2.
Quick unit check — always do this. Put the units into the formula: (N m² C−2) × (C) × (C) ÷ (m²). The C² on top cancels the C−2, the m² on top cancels the m² underneath, and you are left with N. Good — a force. If your units do not collapse to newtons, you have written the formula wrongly. Make this a habit; it catches more errors than re-reading ever will.
The vector form. The scalar formula gives you a size but not a direction, and directions are where marks are lost. Write r21 for the vector pointing from charge 1 to charge 2, and let its unit vector be r̂21. Then the force on charge 2 due to charge 1 is
F21 = (1 ÷ 4πε0) × (q1q2 ÷ r²) × r̂21
This one line is beautifully self-correcting, and here is the trick that makes it worth learning. Put the charges in with their signs:
- If q1 and q2 have the same sign, the product q1q2 is positive, so F21 points along r̂21 — that is, away from charge 1. Repulsion.
- If they have opposite signs, q1q2 is negative, so F21 points opposite to r̂21 — back towards charge 1. Attraction.
You never have to remember “like repels, unlike attracts” as a separate rule; the algebra tells you. And because r̂12 = −r̂21, the same equation immediately gives F12 = −F21, which is Newton’s third law appearing on its own. Nice.
It applies to point charges (or charged bodies small compared with their separation) that are at rest. For moving charges there are extra magnetic effects, and for large overlapping bodies you must integrate over the distribution. A uniformly charged sphere is a happy exception: from outside, it behaves exactly like a point charge sitting at its centre.
Step 1 — SI units first, always. q1 = 2 × 10−6 C, q2 = 3 × 10−6 C, r = 30 cm = 0.30 m.
Step 2 — substitute.
F = (8.99 × 109) × (2 × 10−6)(3 × 10−6) ÷ (0.30)²
Step 3 — numerator. (2 × 10−6)(3 × 10−6) = 6 × 10−12 C².
(8.99 × 109)(6 × 10−12) = 5.394 × 10−2 N m².
Step 4 — denominator. (0.30)² = 0.090 m².
Step 5 — divide. F = (5.394 × 10−2) ÷ 0.090 = 0.599 N ≈ 0.60 N.
Both charges are positive, so the force is repulsive, directed along the line joining them.
Common slip to avoid: forgetting to square the distance, or squaring 30 instead of 0.30. Writing “r = 0.30 m” on its own line before substituting removes both risks.
Step 1 — write both forces. Both go as 1 ÷ r², so the separation will cancel — the ratio is the same at every distance. That is the key insight.
Fe = k e² ÷ r² and Fg = G mp² ÷ r²
Step 2 — take the ratio.
Fe ÷ Fg = k e² ÷ (G mp²)
Step 3 — top. e² = (1.6 × 10−19)² = 2.56 × 10−38 C².
k e² = (8.99 × 109)(2.56 × 10−38) = 2.301 × 10−28.
Step 4 — bottom. mp² = (1.673 × 10−27)² = 2.799 × 10−54 kg².
G mp² = (6.674 × 10−11)(2.799 × 10−54) = 1.868 × 10−64.
Step 5 — divide.
Fe ÷ Fg = (2.301 × 10−28) ÷ (1.868 × 10−64) ≈ 1.23 × 1036
What this means: the electric force between two protons is about a million million million million million million times stronger than their gravitational attraction. This is why we cheerfully ignore gravity in every electrostatics problem in this chapter. (Repeat the calculation with electrons and the ratio is about 4.2 × 1042, even more lopsided, because electrons are far lighter.)
Permittivity and the Effect of a Medium
So far every calculation assumed the charges sat in vacuum (air is close enough). What happens if you dunk the same two charges in oil, or in water? The force gets weaker. Always weaker, never stronger.
Here is the picture. An insulating medium — a dielectric — is full of molecules whose electron clouds can shift slightly. Place a positive charge in the middle and every nearby molecule stretches a little, turning its negative end towards the charge. The medium builds up a thin sheath of opposite charge around each of your charges, and that sheath partially screens them from each other. Less “visible” charge means less force.
We package this into a single number. Replace ε0 by the permittivity of the medium, ε, and define the dielectric constant (also called relative permittivity) as
K = εr = ε ÷ ε0
Then Coulomb’s law in a medium reads
Fmedium = (1 ÷ 4πε0K) × q1q2 ÷ r² = Fvacuum ÷ K
Because K is a ratio of two permittivities, it has no units — a favourite one-mark question. For vacuum K = 1 exactly; for air K ≈ 1.0006, which is why we treat air as vacuum. For water K ≈ 80, which is enormous, and explains why salt dissolves so readily: water weakens the attraction between Na+ and Cl− ions by a factor of about 80, so thermal jostling can pull them apart.
(a) We already have Fair = 0.599 N.
F = Fair ÷ K = 0.599 ÷ 5 = 0.120 N
(b) F = 0.599 ÷ 80 = 7.49 × 10−3 N — the water has knocked the force down to under a hundredth of a newton.
(c) We need k q1q2 ÷ (K r′²) = k q1q2 ÷ r², so K r′² = r², giving
r′ = r ÷ √K = 0.30 ÷ √5 = 0.30 ÷ 2.236 = 0.134 m ≈ 13.4 cm
So to feel the same push inside the medium, the charges must be brought about 13.4 cm apart instead of 30 cm.
Why it works: K sits in the denominator alongside r², so a factor of K in permittivity can always be traded for a factor of √K in distance.
Every real dielectric has K greater than 1. If your answer says the force increased in a medium, you have divided the wrong way round. A quick sanity line — “medium weakens the force” — before you compute will save you the mark.
Forces Between Multiple Charges: The Superposition Principle
Real problems rarely have exactly two charges. So what happens when charge q1 is surrounded by q2, q3, q4 and so on? The answer is delightfully simple, and it is an experimental fact rather than something you can derive:
The total force on any one charge is the vector sum of the forces that each of the other charges would exert on it on its own. The presence of a third charge does not alter the force between the first two.
F1 = F12 + F13 + F14 + …
Read that second sentence again, because it is what makes physics tractable. Charges do not “gang up” or interfere with one another’s interaction. Each pair minds its own business, and you simply add the results as vectors at the end.
The method that never fails. For any multi-charge question, follow this exact sequence:
- Draw a clear diagram and mark the charge you are asked about.
- Draw an arrow for the force from each other charge, pointing correctly (towards for attraction, away for repulsion). Do this before touching your calculator.
- Find the magnitude of each force with Coulomb’s law.
- Resolve into x and y components, add them separately.
- Recombine: F = √(Fx² + Fy²), and give the direction.
Steps 1 and 2 are where the marks are won. A student with a good diagram and a wrong arithmetic slip usually scores more than one with perfect arithmetic and no diagram.
Step 1 — reason out the region first. Both charges are positive, so both push Q away. The two pushes can only cancel if they point in opposite directions, which happens only between the charges. (Outside, both pushes point the same way and can never cancel.)
Step 2 — set up coordinates. Put q1 at x = 0 and q2 at x = 0.50 m. Let Q sit at distance x from q1, so it is (0.50 − x) from q2.
Step 3 — balance the magnitudes.
k q1Q ÷ x² = k q2Q ÷ (0.50 − x)²
k and Q cancel from both sides — note that the answer does not depend on Q at all, nor on its sign:
9 ÷ x² = 4 ÷ (0.50 − x)²
Step 4 — take square roots (both sides positive, so this is safe and avoids a quadratic):
3 ÷ x = 2 ÷ (0.50 − x)
3(0.50 − x) = 2x → 1.50 − 3x = 2x → 5x = 1.50
x = 0.30 m = 30 cm from the 9 μC charge (and therefore 20 cm from the 4 μC charge).
Step 5 — verify. 9 ÷ (0.30)² = 9 ÷ 0.09 = 100. And 4 ÷ (0.20)² = 4 ÷ 0.04 = 100. Equal. ✓
Sense check: the null point is closer to the smaller charge. That has to be true — the weaker charge needs you to come nearer before it can match the stronger one.
Step 1 — each individual force. Both other charges are at the same distance a = 0.10 m and have the same magnitude, so both forces have the same size:
F = k q² ÷ a² = (8.99 × 109)(2 × 10−6)² ÷ (0.10)²
(2 × 10−6)² = 4 × 10−12 C²
(8.99 × 109)(4 × 10−12) = 3.596 × 10−2
F = (3.596 × 10−2) ÷ 0.010 = 3.596 N each
Step 2 — the angle between them. All charges are positive, so both forces on our chosen corner point away from the other two corners. Those two directions are separated by the interior angle of the triangle, 60°.
Step 3 — add the two vectors. For two equal vectors of size F at angle θ:
Fnet = √(F² + F² + 2F² cos θ) = F √(2 + 2 cos 60°) = F √(2 + 1) = √3 F
Step 4 — numbers. Fnet = 1.732 × 3.596 = 6.23 N
Direction: because the two forces are equal, the resultant bisects the 60° angle — it points radially outward from the centroid of the triangle, along the line from the centre through our corner.
Worth remembering: for three equal charges on an equilateral triangle, the answer is always √3 × (the force from one). Recognising the shortcut saves two minutes in the exam.
Step 1 — identify the geometry. Call our charge A, the two adjacent corners B and D (each a distance a = 0.10 m away), and the opposite corner C (a distance a√2 = 0.1414 m away along the diagonal).
Step 2 — forces from the two adjacent charges.
F1 = k q² ÷ a² = (8.99 × 109)(1 × 10−12) ÷ (0.010) = 0.899 N each
These two forces are perpendicular to each other (along the two sides of the square, pointing away from B and away from D).
Step 3 — combine them. Two equal perpendicular vectors give
√(F1² + F1²) = √2 F1 = 1.414 × 0.899 = 1.271 N, directed along the diagonal AC produced, i.e. away from C.
Step 4 — force from the diagonal charge. Distance is a√2, so the distance squared is 2a²:
F2 = k q² ÷ (2a²) = 0.899 ÷ 2 = 0.4495 N
This also points away from C — along the very same diagonal.
Step 5 — add along the diagonal. Everything is now collinear, so simple addition:
Fnet = 1.271 + 0.4495 = 1.72 N, directed outward along the diagonal, away from the centre of the square.
General result you can quote: Fnet = (k q² ÷ a²)(√2 + 0.5) = 1.914 × (k q² ÷ a²).
Why it works: the clever move is Step 3. By combining the two side-forces first, you discover they point along the diagonal — the same line as the third force. After that there is no vector work left at all.
In Example 9 the diagonal distance is a√2, so r² = 2a², not a²√2. Students who miss this get F2 = 0.636 N instead of 0.4495 N. Write the distance first, square it on its own line, then substitute.
Electric Field and the Field of a Point Charge
Coulomb’s law has one philosophical wrinkle that bothered physicists for a century: how does charge A know that charge B is over there? They are not touching. Action at a distance across empty space feels like magic.
The resolution is the field. We say that a charge changes the space around it — it fills that space with an invisible condition. Any other charge placed there feels a force because of the local condition of the space at its own position, not because of a message sent instantly across the gap. It is a change of viewpoint, but a powerful one, and it is how all of modern physics is written.
Definition. The electric field at a point is the force per unit positive charge placed at that point:
E = F ÷ q0
Here q0 is a test charge — imagined to be so small that it does not disturb the charges creating the field. Strictly, E is the limit of F/q0 as q0 tends to zero. The field is a vector, and its SI unit is the newton per coulomb, N C−1 (you will meet the equivalent unit volt per metre, V m−1, in the next chapter).
Field of a single point charge. Put a test charge q0 a distance r from a point charge q. Coulomb’s law gives F = k q q0 ÷ r². Divide by q0 and the test charge disappears:
E = (1 ÷ 4πε0) × q ÷ r²
The direction is radially outward for a positive source charge and radially inward for a negative one. Notice how satisfying this is: the field belongs to the source charge alone. It exists whether or not there is anything there to feel it.
Unit check. (N m² C−2)(C) ÷ (m²) = N C−1. ✓
Superposition works for fields exactly as it did for forces: E = E1 + E2 + E3 + …, added as vectors. And once you know E at a point, the force on any charge q placed there is simply F = qE. For a positive q the force is along E; for a negative q it is opposite to E.
E at a point depends only on the source charge, and points away from a positive source.
F = qE on a placed charge depends on the sign of that placed charge. An electron in a rightward field feels a leftward force.
(a) Step 1 — SI units. q = 4 × 10−6 C, r = 0.20 m, r² = 0.040 m².
Step 2 — substitute.
E = k q ÷ r² = (8.99 × 109)(4 × 10−6) ÷ 0.040
Numerator: 3.596 × 104
E = (3.596 × 104) ÷ 0.040 = 8.99 × 105 N C−1, directed radially away from the charge.
(b) Step 3 — use F = qE. q = −2 nC = −2 × 10−9 C.
|F| = (2 × 10−9)(8.99 × 105) = 1.80 × 10−3 N
Because the charge is negative, the force points opposite to E, i.e. back towards the +4 μC charge. Attraction, as expected for unlike charges.
Step 1 — force on the electron.
F = eE = (1.6 × 10−19)(1.00 × 103) = 1.6 × 10−16 N
Step 2 — Newton’s second law.
a = F ÷ me = (1.6 × 10−16) ÷ (9.11 × 10−31)
1.6 ÷ 9.11 = 0.1756, and 10−16 ÷ 10−31 = 1015
a = 0.1756 × 1015 = 1.76 × 1014 m s−2, directed opposite to E.
Perspective: that is roughly 1013 times the acceleration due to gravity. This is exactly why gravity is ignored for charged particles in fields — and it is also the principle behind the electron gun in an old television tube.
Step 1 — locate the region. Both fields point away from their own charge. They can only cancel between the two charges, where they point in opposite directions.
Step 2 — set up. Let the point be x from the 16 μC charge, so (0.60 − x) from the 4 μC charge.
k(16 × 10−6) ÷ x² = k(4 × 10−6) ÷ (0.60 − x)²
Step 3 — cancel and square-root.
16 ÷ x² = 4 ÷ (0.60 − x)² → 4 ÷ x = 2 ÷ (0.60 − x)
4(0.60 − x) = 2x → 2.40 − 4x = 2x → x = 0.40 m = 40 cm from the 16 μC charge.
Step 4 — verify. 16 ÷ (0.40)² = 16 ÷ 0.16 = 100; 4 ÷ (0.20)² = 4 ÷ 0.04 = 100. Equal. ✓
Contrast with Example 7: exactly the same algebra. A “zero force” question and a “zero field” question are the same problem, because F = qE and the q cancels out.
Electric Field Lines and Their Properties
An electric field is a vector at every single point in space, which is impossible to draw. Michael Faraday’s solution was to draw a few continuous curves such that the tangent at any point gives the field direction there, and the crowding of the curves shows the field strength. These are electric field lines.
The picture to hold in your head: field lines are like the streamlines of water flowing out of a fountain. Where the streamlines bunch together, the flow is fast; where they spread apart, it is slow.
The properties, which are examined almost every year:
- Field lines start on positive charges and end on negative charges. If there is no negative charge nearby, they run off to infinity; if there is no positive charge nearby, they come in from infinity.
- The tangent at any point on a line gives the direction of E at that point.
- The density of lines (number crossing unit area held perpendicular to them) is proportional to the magnitude of E. Crowded lines mean a strong field.
- Field lines are continuous smooth curves with no breaks in a charge-free region.
- Two field lines can never cross. If they did, there would be two tangents at the crossing point, meaning the field has two directions there at once — which is meaningless.
- Electrostatic field lines do not form closed loops. (Reason: the electrostatic field is conservative. Going round a closed loop, the work done must be zero, which a closed field line could never satisfy.)
- They are always perpendicular to the surface of a conductor in electrostatic equilibrium, and there are no lines inside a conductor, because the field inside is zero.
- Lines never pass through a charge; they begin or end there.
The standard pictures you should be able to sketch: a single positive charge (straight lines radiating outward, evenly spaced); a single negative charge (the same, but pointing inward); two equal and opposite charges — a dipole (curved lines leaving the positive and looping round into the negative, densest in the gap between them); two equal positive charges (lines pushing away from each other with a clear neutral point at the midpoint where no line passes, because the two fields cancel exactly); and a uniform field (parallel, equally spaced straight lines, as found between two oppositely charged parallel plates).
They are not, in general. A charge released from rest does start moving along the line, but as soon as it has velocity it carries momentum, and on a curved field line its trajectory peels away from the line. The two coincide only when the field lines are straight.
“Why can two electric field lines never intersect?” The full-mark answer is one sentence: at the point of intersection there would be two tangents, hence two directions of the electric field at the same point, which is impossible. Learn it exactly like that.
Electric Dipole and Dipole Moment
An electric dipole is a pair of equal and opposite point charges, +q and −q, held a small distance apart. Call that separation 2a. This is not an artificial construction — a water molecule is a dipole, HCl is a dipole, and any neutral atom in an external field becomes a temporary one.
The total charge of a dipole is zero. So from very far away it looks like nothing at all. But close up its two ends do not coincide, and that separation produces a real field. To capture “how strong a dipole is” we define the electric dipole moment:
p = q × 2a
It is a vector, and by convention it points from the negative charge towards the positive charge. Its magnitude is p = q(2a), where q is the magnitude of either charge (not their sum — a very common slip). The SI unit is the coulomb metre, C m.
Two traps in one place. (1) p points from −q to +q. This is the opposite of the direction of the electric field between the charges, which runs from + to −. (2) In p = q(2a), take q as the magnitude of one charge. For ±5 μC separated by 2 cm, p = (5 × 10−6)(0.02) = 1 × 10−7 C m — not 2 × 10−7.
What “short dipole” means. Nearly every board question says “for a short dipole” or “r >> a”. This is permission to throw away the small quantity a wherever it appears alongside r. It is what turns the exact, ugly expressions of the next section into the two clean formulas you will actually memorise. An ideal or point dipole is the mathematical limit where 2a tends to zero and q grows so that p stays fixed.
Field of a Dipole: Axial and Equatorial Lines
Two special lines matter. The axial line is the line through both charges, extended. The equatorial line is the perpendicular bisector of the dipole. Board questions ask for the derivation on either line, so let us do both properly. Set the centre of the dipole at O, with −q and +q a distance a on each side.
Derivation 1 — field on the axial line. Take a point P on the axis, a distance r from the centre O, on the +q side.
- P is a distance (r − a) from +q, and (r + a) from −q.
- Field due to +q, pointing away from the dipole (in the direction of p): E+ = kq ÷ (r − a)²
- Field due to −q, pointing back towards the dipole (opposite to p): E− = kq ÷ (r + a)²
- Since P is nearer to +q, E+ wins. The net field points along p:
Eaxial = kq [ 1 ÷ (r − a)² − 1 ÷ (r + a)² ] - Put over a common denominator. The numerator is (r + a)² − (r − a)² = 4ar, and the denominator is (r² − a²)²:
Eaxial = kq × 4ar ÷ (r² − a²)² - Recognise q × 2a = p, so 4aqr = 2pr:
Eaxial = 2kpr ÷ (r² − a²)² (exact) - For a short dipole, r >> a, so a² is negligible next to r² and (r²)² = r4:
Eaxial = 2kp ÷ r³ = (1 ÷ 4πε0) × 2p ÷ r³, directed parallel to p.
Derivation 2 — field on the equatorial line. Take a point P on the perpendicular bisector, a distance r from the centre O.
- By symmetry, P is the same distance from both charges: √(r² + a²).
- Each charge therefore produces a field of the same magnitude:
E+ = E− = kq ÷ (r² + a²) - E+ points away from +q; E− points towards −q. Resolve both along and perpendicular to the dipole axis.
- The components perpendicular to the axis are equal and opposite, so they cancel. This is the key step — say it explicitly in the exam.
- The components parallel to the axis both point from +q towards −q, i.e. antiparallel to p, so they add. Each contributes a factor cos θ, where θ is the angle at P, and from the geometry cos θ = a ÷ √(r² + a²).
- Eequatorial = 2 × [ kq ÷ (r² + a²) ] × [ a ÷ √(r² + a²) ] = k(2aq) ÷ (r² + a²)3/2
Eequatorial = kp ÷ (r² + a²)3/2 (exact) - For a short dipole, r >> a:
Eequatorial = kp ÷ r³ = (1 ÷ 4πε0) × p ÷ r³, directed antiparallel to p.
At the same distance r from a short dipole:
Eaxial = 2kp ÷ r³, along p
Eequatorial = kp ÷ r³, opposite to p
So Eaxial = 2 × Eequatorial, and the two point in opposite senses.
Also note the 1 ÷ r³ dependence — a dipole’s field dies away faster than a point charge’s 1 ÷ r², because from far off the two charges nearly cancel.
(a) Step 1 — SI units. q = 2 × 10−9 C, 2a = 2.0 cm = 0.020 m.
p = q × 2a = (2 × 10−9)(0.020) = 4.0 × 10−11 C m, directed from −q to +q.
(b) Step 2 — check the short-dipole condition. r = 0.20 m and a = 0.010 m, so r is twenty times a. Comfortably short. Use E = 2kp ÷ r³.
Step 3 — cube the distance. r³ = (0.20)³ = 8.0 × 10−3 m³.
Step 4 — substitute.
E = 2(8.99 × 109)(4.0 × 10−11) ÷ (8.0 × 10−3)
Numerator: 2 × 8.99 × 4.0 = 71.92, and 109 × 10−11 = 10−2, giving 0.7192.
E = 0.7192 ÷ (8.0 × 10−3) = 89.9 N C−1 ≈ 90 N C−1, parallel to p.
How good is the approximation? The exact formula 2kpr ÷ (r² − a²)² gives 90.4 N C−1. The short-dipole shortcut is off by about 0.5% — well inside board tolerance, and it saves you a page of arithmetic.
Step 1 — use the equatorial formula.
The same short-dipole condition still holds (r = 0.20 m is twenty times a = 0.010 m, so r >> a) — say so in the exam, because the marking scheme gives credit for justifying the approximation.
E = kp ÷ r³ = (8.99 × 109)(4.0 × 10−11) ÷ (8.0 × 10−3)
Numerator: 8.99 × 4.0 = 35.96, times 10−2 = 0.3596
E = 0.3596 ÷ (8.0 × 10−3) = 44.95 N C−1 ≈ 45 N C−1
Direction: antiparallel to p, i.e. pointing from the +q end back towards the −q end.
Step 2 — the ratio.
Eaxial ÷ Eequatorial = 89.9 ÷ 44.95 = 2
Why it works: the ratio is exactly 2 for any short dipole at any distance, because the r³ and the kp are identical in both formulas — only the leading factor differs. If a question gives you one field and asks for the other at the same distance, you just double or halve. No calculator needed.
Torque on a Dipole in a Uniform Field
Now place a dipole inside a uniform external field E, with the dipole axis making an angle θ with the field. What happens to it?
First, the net force. The +q end feels a force qE along E. The −q end feels a force qE opposite to E. Equal magnitudes, opposite directions:
Fnet = qE − qE = 0
So a dipole in a uniform field does not translate. Its centre of mass stays put.
But it does rotate. The two forces are equal, opposite, and act along different lines — that is precisely the definition of a couple, and a couple produces pure rotation.
Derivation of the torque. Torque of a couple = (either force) × (perpendicular distance between the two lines of action).
- Either force has magnitude qE.
- The two charges are 2a apart, but the two lines of action are separated by the perpendicular distance 2a sin θ. Draw this — it is the only geometry step, and it is where marks are lost.
- τ = (qE)(2a sin θ) = (q × 2a) E sin θ
- Since q × 2a = p:
τ = pE sin θ, and in vector form τ = p × E.
Reading the formula. The torque is maximum when θ = 90° (dipole perpendicular to the field), where τmax = pE. It is zero when θ = 0° or 180°, but those two zeros are not the same: at θ = 0° the dipole is aligned with the field and the equilibrium is stable (nudge it and the torque pushes it back); at θ = 180° it is anti-aligned and the equilibrium is unstable (the tiniest nudge and it flips right round). A compass needle settling to north is the magnetic version of the same story.
Unit check. [p][E] = (C m)(N C−1) = N m. That is a torque. ✓ A neat side-effect: this also shows that C m × N C−1 works out, so the unit of dipole moment really must be the coulomb metre.
Examiners like to ask what changes if the field is non-uniform. Answer: the two charges now sit in different field strengths, so qE1 ≠ qE2, the forces no longer cancel, and the dipole experiences a net force as well as a torque — it both rotates and drifts. That is why a charged comb attracts neutral paper bits: it polarises them into dipoles and then pulls them in with its non-uniform field.
(a) Step 1 — find p.
p = q × 2a = (2 × 10−6)(0.030) = 6.0 × 10−8 C m
Step 2 — apply τ = pE sin θ, with sin 30° = 0.5:
τ = (6.0 × 10−8)(5.0 × 105)(0.5)
(6.0 × 10−8)(5.0 × 105) = 3.0 × 10−2
τ = (3.0 × 10−2)(0.5) = 1.5 × 10−2 N m
(b) Maximum torque occurs at θ = 90°, where sin θ = 1:
τmax = pE = 3.0 × 10−2 N m
(c) The field is uniform, so the two forces are equal and opposite: net force = 0. The dipole rotates but does not move off.
Continuous Charge Distributions: λ, σ and ρ
So far every charge has been a neat dot. Real objects — a charged wire, a metal plate, a plastic ball — carry charge spread over a length, an area or a volume, with billions of elementary charges packed so closely that treating them one at a time is hopeless. So we smooth them out and speak of a continuous charge distribution, described by a density.
| Distribution | Symbol | Definition | SI unit | Total charge |
|---|---|---|---|---|
| Linear (a wire, a rod) | λ | charge per unit length, λ = Δq ÷ Δl | C m−1 | Q = λL (if uniform) |
| Surface (a sheet, a shell) | σ | charge per unit area, σ = Δq ÷ ΔS | C m−2 | Q = σA (if uniform) |
| Volume (a solid body) | ρ | charge per unit volume, ρ = Δq ÷ ΔV | C m−3 | Q = ρV (if uniform) |
How you use them. Chop the object into elements so small that each looks like a point charge — Δq = λΔl, or σΔS, or ρΔV. Each element contributes a field kΔq ÷ r² at your chosen point. Then add up all the contributions as vectors, which in the limit becomes an integral. In Class 12 you almost never perform that integral: Gauss’s theorem does the hard work for you, which is exactly why it is coming up next.
(a) A straight wire of length 2.0 m with uniform λ = 5.0 μC m−1.
(b) A flat plate of area 0.25 m² with uniform σ = 4.0 μC m−2.
(c) A solid sphere of radius 10 cm with uniform ρ = 3.0 μC m−3.
(a) Q = λL = (5.0 × 10−6)(2.0) = 1.0 × 10−5 C = 10 μC
(b) Q = σA = (4.0 × 10−6)(0.25) = 1.0 × 10−6 C = 1.0 μC
(c) Step 1 — volume of the sphere.
V = (4 ÷ 3)πR³ = (4 ÷ 3)(3.1416)(0.10)³ = (4 ÷ 3)(3.1416)(1.0 × 10−3) = 4.189 × 10−3 m³
Step 2 — multiply.
Q = ρV = (3.0 × 10−6)(4.189 × 10−3) = 1.26 × 10−8 C = 12.6 nC
Watch the unit conversions: in (c), 10 cm must become 0.10 m before cubing. Cubing 10 instead of 0.10 makes the answer wrong by a factor of a million.
Electric Flux
Think of the electric field as something flowing, like wind. Hold a rectangular hoop up in the wind. How much air passes through it? That depends on the wind speed, the size of the hoop, and crucially the orientation — hold the hoop face-on and you catch a lot; turn it edge-on and you catch nothing at all. Electric flux is exactly this idea applied to the electric field.
To handle orientation properly we give every surface element an area vector ΔS: its magnitude is the area, and its direction is along the outward normal to the surface. Then the flux through that element is
Δφ = E · ΔS = E ΔS cos θ
where θ is the angle between E and the normal — not the angle with the surface itself. For a whole surface you add up all the elements, written as the integral φ = ∫ E · dS.
Flux is a scalar, even though it is built from two vectors, because the dot product of two vectors is a scalar. But it carries a sign: flux is positive where field lines leave the surface (θ < 90°) and negative where they enter it (θ > 90°). Its SI unit is N m² C−1 (equivalently V m).
Three orientations are worth memorising:
- Surface perpendicular to the field (normal parallel to E, θ = 0°): φ = EA, the maximum.
- Surface at an angle, normal making θ with E: φ = EA cos θ.
- Surface parallel to the field (normal perpendicular to E, θ = 90°): φ = 0. The lines skim past without crossing.
“The plane makes 30° with the field” is not the same as “the normal makes 30° with the field”. If the plane is at 30° to E, the normal is at 60°, so φ = EA cos 60°, not EA cos 30°. Whenever a question gives you an angle, write down explicitly whether it is measured to the plane or to the normal before you substitute.
Step 1 — area. A = (0.10)² = 1.0 × 10−2 m².
(a) θ = 0°, cos 0° = 1.
φ = EA = (3.0 × 103)(1.0 × 10−2) = 30 N m² C−1
(b) θ = 60°, cos 60° = 0.5.
φ = EA cos 60° = 30 × 0.5 = 15 N m² C−1
(c) If the plane contains the x-axis, the field lies in the surface, so the normal is perpendicular to E, θ = 90°, cos 90° = 0:
φ = 0
Why it works: only the component of E along the normal pushes anything “through” the surface. The component lying in the plane simply slides across it and contributes nothing.
Gauss’s Theorem
This is the most powerful single statement in the chapter, and the one the board loves most. It connects the flux through any closed surface to the charge inside it — and to nothing else.
The total electric flux through any closed surface is 1 ÷ ε0 times the net charge enclosed by that surface.
φ = ∫ E · dS = qenclosed ÷ ε0
The imaginary closed surface you choose is called a Gaussian surface.
Four consequences you must be able to state:
- The flux depends only on the net enclosed charge — not on its position inside, not on its shape, not on how it is spread out.
- The flux does not depend on the size or shape of the Gaussian surface. A charge at the centre of a tiny sphere and the same charge inside a vast cube give identical flux.
- Charges outside the surface contribute zero net flux. Their field lines enter one side and leave the other, so the negative and positive contributions cancel exactly. (Careful: those outside charges do affect the field E at points on the surface. They just do not change the total flux.)
- If the net enclosed charge is zero, the total flux is zero — but the field need not be zero anywhere.
A quick plausibility argument. Put a point charge q at the centre of a sphere of radius r. By symmetry E has the same magnitude everywhere on the sphere and points radially outward, parallel to every area element. So the flux is just E times the total area:
φ = E × 4πr² = [ q ÷ (4πε0r²) ] × 4πr² = q ÷ ε0
Every r has cancelled. The result is independent of the radius — which is Gauss’s theorem appearing before your eyes. It works because the field falls as 1 ÷ r² while the area grows as r², and the two effects exactly annul. If Coulomb’s law had any other exponent, Gauss’s theorem would be false.
The real skill is choosing the Gaussian surface. Gauss’s theorem is always true, but only useful when symmetry lets you pull E outside the integral. Match the surface to the symmetry of the charge: a sphere for a point charge or a spherical shell, a cylinder for a long straight wire, a cylinder or box straddling the plane for an infinite sheet. Get this choice right and the derivation is three lines long.
(a) Step 1 — Gauss’s theorem, straight away. Notice that the side length is not needed at all.
φ = q ÷ ε0 = (2.0 × 10−6) ÷ (8.854 × 10−12)
2.0 ÷ 8.854 = 0.2259, and 10−6 ÷ 10−12 = 106
φ = 0.2259 × 106 = 2.26 × 105 N m² C−1
(b) Step 2 — use symmetry. The charge sits at the centre, so all six faces are equivalent and each takes one sixth:
φface = (2.26 × 105) ÷ 6 = 3.76 × 104 N m² C−1
(c) Nothing changes. Flux depends only on the enclosed charge, not the size of the surface. Both answers stay the same.
(d) The total flux is still 2.26 × 105 N m² C−1, because the same charge is still enclosed. But the per-face answer is no longer valid — the symmetry is broken, and the nearer faces now receive more flux than the farther ones.
Why the one-sixth trick is allowed: only because of the perfect symmetry of a centred charge in a cube. Take that symmetry away and the division fails. Examiners test this exact distinction.
Application 1: Infinitely Long Straight Charged Wire
An infinitely long straight wire carries a uniform linear charge density λ. Find the field at a perpendicular distance r from it.
Step 1 — use the symmetry. The wire looks the same from every direction around it and from every point along it. So E can only point radially outward (perpendicular to the wire), and its magnitude can depend only on r.
Step 2 — choose the Gaussian surface. A coaxial cylinder of radius r and length l, with the wire running along its axis. It has three parts: the curved side and the two flat end caps.
Step 3 — flux through each part.
- Two end caps: their outward normals point along the wire, while E points radially outward — perpendicular to them. So θ = 90° and the flux through each cap is zero.
- Curved surface: here the outward normal is radial, exactly parallel to E, and E has the same value everywhere on it (same r). So the flux is E × (curved area) = E × 2πrl.
Step 4 — total flux. φ = E(2πrl) + 0 + 0 = E(2πrl)
Step 5 — enclosed charge. The cylinder encloses a length l of wire, so qenc = λl.
Step 6 — apply Gauss’s theorem.
E(2πrl) = λl ÷ ε0
The length l cancels from both sides — which is reassuring, because our choice of l was arbitrary:
E = λ ÷ (2πε0r) = 2kλ ÷ r
Direction: radially outward if λ is positive, radially inward if negative. Note the 1 ÷ r dependence — slower than a point charge’s 1 ÷ r², because as you move away you “see” more of the wire, which partly compensates.
Step 1 — use E = 2kλ ÷ r. λ = 4.0 × 10−6 C m−1, r = 0.20 m.
Step 2 — numerator. 2kλ = 2(8.99 × 109)(4.0 × 10−6) = 7.192 × 104
Step 3 — divide. E = (7.192 × 104) ÷ 0.20 = 3.60 × 105 N C−1, radially outward.
At 40 cm: because E goes as 1 ÷ r, doubling the distance simply halves the field:
E = (3.60 × 105) ÷ 2 = 1.80 × 105 N C−1
Contrast this with a point charge, where doubling the distance would have cut the field to a quarter. Recognising which power law applies is worth a mark on its own.
Application 2: Uniformly Charged Infinite Plane Sheet
An infinite plane sheet carries a uniform surface charge density σ. Find the field at a distance from it.
Step 1 — symmetry. Every point on the sheet looks like every other point, so E must be perpendicular to the sheet everywhere (any sideways component would have no reason to choose one direction over another). By reflection symmetry the field points away from the sheet on both sides with the same magnitude, and can depend at most on the perpendicular distance.
Step 2 — Gaussian surface. A small cylinder (a pillbox) pushed through the sheet, with its flat circular ends of area A parallel to the sheet and at equal distances on either side.
Step 3 — flux through each part.
- Curved side: its normal lies in the plane of the sheet, perpendicular to E. Flux = 0.
- Each flat end: normal is parallel to E (pointing outward on both sides), so flux = EA through each. Two ends give 2EA.
Step 4 — enclosed charge. The pillbox cuts out an area A of the sheet, so qenc = σA.
Step 5 — Gauss’s theorem.
2EA = σA ÷ ε0 → A cancels:
E = σ ÷ (2ε0)
There is no r in the answer. The field of an infinite charged sheet is the same at every distance — it is a genuinely uniform field. Moving further away means the sheet’s contributions arrive weaker but from a wider spread, and for an infinite sheet these balance perfectly. In practice this holds close to a large finite sheet, where the edges are far away.
Two useful extensions. Bring up a second, parallel sheet with charge density −σ. In the region between the sheets the two fields point the same way and add, giving E = σ ÷ ε0; outside the pair they oppose and cancel, giving E = 0. That is the parallel-plate capacitor, which you will meet properly in the next chapter. And for the surface of a charged conductor, the field just outside is E = σ ÷ ε0 — twice the sheet value, because the field exists only on the outside (it is zero inside the conductor), so one pillbox end contributes instead of two.
(a) Step 1 — substitute into E = σ ÷ 2ε0.
2ε0 = 2(8.854 × 10−12) = 1.7708 × 10−11
E = (2.0 × 10−6) ÷ (1.7708 × 10−11)
2.0 ÷ 1.7708 = 1.129, and 10−6 ÷ 10−11 = 105
E = 1.13 × 105 N C−1, directed away from the sheet on both sides (σ is positive). This value holds at 1 cm, at 1 m — anywhere near the sheet.
(b) Between the sheets: the fields add.
E = σ ÷ ε0 = 2 × (1.13 × 105) = 2.26 × 105 N C−1, pointing from the positive sheet to the negative one.
Outside the pair: the two fields are equal and opposite, so E = 0.
Application 3: Uniformly Charged Thin Spherical Shell
This is the derivation that appears most often, because it has two parts — outside and inside — and the inside answer surprises people. A thin spherical shell of radius R carries a total charge Q spread uniformly over its surface, so σ = Q ÷ (4πR²).
Symmetry first. The charge distribution is perfectly spherically symmetric, so E must be radial and its magnitude can depend only on the distance r from the centre. In both cases we choose a concentric sphere of radius r as the Gaussian surface, on which E is constant and everywhere parallel to the outward normal. So in both cases
φ = E × 4πr²
Case 1 — outside the shell (r > R). The Gaussian sphere encloses the entire shell, so qenc = Q. Gauss’s theorem gives
E × 4πr² = Q ÷ ε0
E = Q ÷ (4πε0r²) = kQ ÷ r²
This is exactly the field of a point charge Q sitting at the centre. From outside, a uniformly charged shell is indistinguishable from a point charge at its centre — a beautiful result, and the electrostatic twin of the shell theorem in gravitation.
Case 2 — on the surface (r = R). Setting r = R:
E = kQ ÷ R², and substituting Q = σ(4πR²) gives the equivalent form E = σ ÷ ε0.
Case 3 — inside the shell (r < R). Here is the interesting one. Draw the Gaussian sphere inside the shell. All the charge sits on the shell’s surface, at radius R, which is outside our Gaussian sphere. Therefore
qenc = 0, so E × 4πr² = 0 ÷ ε0 = 0
E = 0 everywhere inside the shell
Why the inside field is zero — the physical picture. Stand anywhere inside the shell, not necessarily at the centre. Look in one direction and you see a small nearby patch of charge; look the opposite way and you see a much larger patch, but further away. The patch areas grow as the square of the distance, and the field falls as the inverse square of the distance. The two effects cancel exactly, from every direction, at every interior point. This is why a metal box screens out external electric fields — the principle behind electrostatic shielding, and the reason a car is a fairly safe place to be in a lightning storm.
r > R: E = kQ ÷ r² (acts like a point charge at the centre)
r = R: E = kQ ÷ R² = σ ÷ ε0 (the maximum value)
r < R: E = 0
A graph of E against r therefore stays flat at zero from r = 0 up to r = R, jumps to its maximum kQ ÷ R² at the surface, then falls away as 1 ÷ r². Sketching this graph correctly is often worth a mark by itself.
(a) Outside, r = 0.20 m > R. Treat as a point charge at the centre.
E = kQ ÷ r² = (8.99 × 109)(5.0 × 10−6) ÷ (0.20)²
Numerator: 4.495 × 104; denominator: 0.040
E = (4.495 × 104) ÷ 0.040 = 1.12 × 106 N C−1, radially outward.
(b) On the surface, r = R = 0.10 m. r² = 0.010 m²
E = (4.495 × 104) ÷ 0.010 = 4.5 × 106 N C−1
(c) Inside, r = 0.050 m < R. The Gaussian sphere encloses no charge, so
E = 0
(d) Surface charge density.
σ = Q ÷ (4πR²) = (5.0 × 10−6) ÷ [4(3.1416)(0.10)²]
4πR² = 4(3.1416)(0.010) = 0.12566 m²
σ = (5.0 × 10−6) ÷ 0.12566 = 3.98 × 10−5 C m−2
Check: E = σ ÷ ε0 = (3.98 × 10−5) ÷ (8.854 × 10−12) = 4.5 × 106 N C−1. This matches part (b). ✓
Notice the ratio: going from r = R to r = 2R dropped the field by a factor of 4, exactly as 1 ÷ r² requires. Checking a ratio like this takes five seconds and catches most arithmetic errors.
The whole trick: Gauss’s theorem cares only about the charge enclosed by your Gaussian sphere. So the question becomes “what is inside radius r?”
(a) r = 0.050 m. This sphere encloses only the central point charge; the shell’s charge lies outside it.
qenc = 2.0 × 10−6 C
E = kqenc ÷ r² = (8.99 × 109)(2.0 × 10−6) ÷ (0.050)²
Numerator: 1.798 × 104; denominator: 2.5 × 10−3
E = 7.19 × 106 N C−1, radially outward.
(b) r = 0.20 m. Now the sphere encloses both charges.
qenc = 5.0 + 2.0 = 7.0 μC = 7.0 × 10−6 C
E = (8.99 × 109)(7.0 × 10−6) ÷ (0.20)² = (6.293 × 104) ÷ 0.040
E = 1.57 × 106 N C−1, radially outward.
The lesson: “E = 0 inside a shell” is a statement about the shell’s own contribution. Put a charge inside and that charge produces its own field there, undisturbed. Always ask what your Gaussian surface actually encloses — never recite the result blindly.
A thin shell has all its charge on the surface, so E = 0 everywhere inside. That is not the same object as a uniformly charged solid insulating sphere, where charge fills the volume and the internal field grows linearly with r. Read the words “thin spherical shell” carefully — the board always states which one it means, and the CBSE syllabus asks only for the shell.
Formula Sheet at a Glance
Copy this table into the back of your notebook. Read it once before every practice session — not to cram, but so the shapes become familiar. Throughout, k = 1 ÷ (4πε0) = 8.99 × 109 N m² C−2, and ε0 = 8.854 × 10−12 C² N−1 m−2.
| Quantity | Formula | SI unit | Note |
|---|---|---|---|
| Quantisation of charge | q = ne | C | n an integer; e = 1.6 × 10−19 C |
| Coulomb’s law | F = k q1q2 ÷ r² | N | Point charges at rest |
| Force in a medium | F = Fvac ÷ K | N | K = ε ÷ ε0, dimensionless, K ≥ 1 |
| Electric field | E = F ÷ q0; F = qE | N C−1 | Vector |
| Field of a point charge | E = kq ÷ r² | N C−1 | Radial |
| Dipole moment | p = q × 2a | C m | Points from −q to +q |
| Axial field (short dipole) | E = 2kp ÷ r³ | N C−1 | Parallel to p; exact: 2kpr ÷ (r² − a²)² |
| Equatorial field (short dipole) | E = kp ÷ r³ | N C−1 | Antiparallel to p; exact: kp ÷ (r² + a²)3/2 |
| Torque on a dipole | τ = pE sin θ; τ = p × E | N m | Net force is zero in a uniform field |
| Electric flux | φ = EA cos θ | N m² C−1 | θ measured to the normal; scalar |
| Gauss’s theorem | φ = qenc ÷ ε0 | N m² C−1 | Any closed surface |
| Infinite straight wire | E = λ ÷ (2πε0r) = 2kλ ÷ r | N C−1 | Falls as 1 ÷ r |
| Infinite plane sheet | E = σ ÷ (2ε0) | N C−1 | Independent of distance |
| Between two opposite sheets | E = σ ÷ ε0 | N C−1 | Zero outside the pair |
| Thin spherical shell | r > R: E = kQ ÷ r²; r < R: E = 0 | N C−1 | At r = R, E = σ ÷ ε0 |
Point charge: 1 ÷ r². Dipole: 1 ÷ r³. Long wire: 1 ÷ r. Infinite sheet: constant. If you remember only these four shapes, you can reconstruct or sanity-check almost every formula in the chapter, and you will spot a wrong answer instantly.
Practice Worksheet
Ten questions, arranged roughly in increasing difficulty. Do them with a closed book and a blank sheet of paper — that is the only way to find out what you actually know. Give yourself about 45 minutes. Reveal each answer only after you have written something down, even if you are unsure. Take k = 8.99 × 109 N m² C−2, ε0 = 8.854 × 10−12 C² N−1 m−2, e = 1.6 × 10−19 C.
Show Answer
n = q ÷ e = (1.6 × 10−6) ÷ (1.6 × 10−19)
1.6 ÷ 1.6 = 1, and 10−6 ÷ 10−19 = 1013
n = 1.0 × 1013 electrons removed.
Property used: quantisation of charge — any observable charge is an integral multiple of the elementary charge e, i.e. q = ne where n is an integer.
Show Answer
F = k q1q2 ÷ r² = (8.99 × 109)(4.0 × 10−6)(6.0 × 10−6) ÷ 0.16
Numerator: (8.99 × 109)(2.4 × 10−11) = 0.21576
F = 0.21576 ÷ 0.16 = 1.35 N
The charges are unlike, so the force is attractive.
(b) F = Fair ÷ K = 1.35 ÷ 3 = 0.449 N ≈ 0.45 N, still attractive.
(c) The dielectric constant is a ratio of two permittivities, so it has no unit (dimensionless).
Show Answer
q′ = (6.0 + 2.0) ÷ 2 = +4.0 μC on each.
Step 2 — compare the forces. The distance r is unchanged, so F is proportional to the product of the charges.
Original product: 6.0 × 2.0 = 12 (in μC²)
New product: 4.0 × 4.0 = 16
Step 3 — ratio.
Fnew ÷ Fold = 16 ÷ 12 = 4 : 3 ≈ 1.33
Insight: for a fixed total charge, the product q1q2 is largest when the two charges are equal. So sharing always increases the repulsion between two like charges.
Show Answer
kQ ÷ x² = k(4Q) ÷ (0.90 − x)²
Cancel kQ and take square roots: 1 ÷ x = 2 ÷ (0.90 − x)
0.90 − x = 2x → 3x = 0.90 → x = 0.30 m = 30 cm from Q (60 cm from 4Q).
Check: 1 ÷ (0.30)² = 11.11 and 4 ÷ (0.60)² = 11.11. ✓
(b) Yes, but it moves outside the pair, on the far side of the smaller charge. Let it be x beyond Q, so its distance from 4Q is (x + 0.90).
1 ÷ x² = 4 ÷ (x + 0.90)² → (x + 0.90) ÷ x = 2 → x = 0.90 m beyond Q, on the side away from 4Q.
Check: 1 ÷ 0.81 = 1.235 and 4 ÷ (1.80)² = 4 ÷ 3.24 = 1.235. ✓
Rule of thumb: like charges → null point between them; unlike charges → null point outside, beyond the weaker one.
Show Answer
p = q × 2a = (10 × 10−6)(2.0 × 10−3) = 2.0 × 10−8 C m, directed from −q to +q.
(b) τmax = pE (at θ = 90°) = (2.0 × 10−8)(1.0 × 105) = 2.0 × 10−3 N m
(c) τ = pE sin 30° = (2.0 × 10−3)(0.5) = 1.0 × 10−3 N m
(d) The field is uniform, so the forces qE and −qE cancel: net force = 0. The dipole rotates but does not translate.
(e) At θ = 0°, with p aligned along E. At θ = 180° the torque is also zero, but that equilibrium is unstable — the smallest disturbance makes the dipole swing right round.
Show Answer
E = 2kp ÷ r³ = 2(8.99 × 109)(6.0 × 10−8) ÷ 0.027
Numerator: 2 × 8.99 × 6.0 = 107.88, and 109 × 10−8 = 10, so 1078.8
E = 1078.8 ÷ 0.027 = 4.00 × 104 N C−1, parallel to p.
(b) The equatorial field is exactly half the axial field at the same distance:
E = 2.00 × 104 N C−1, antiparallel to p.
(c) On the axial line E is proportional to 1 ÷ r³. To make E one-eighth, r³ must become 8 times bigger, so r must become 3√8 = 2 times bigger:
r = 60 cm
Check: 2kp ÷ (0.60)³ = 1078.8 ÷ 0.216 = 4.99 × 103, which is indeed one-eighth of 4.00 × 104. ✓
Show Answer
φcube = q ÷ (8ε0) = (8.0 × 10−6) ÷ [8(8.854 × 10−12)]
= (1.0 × 10−6) ÷ (8.854 × 10−12) = 1.13 × 105 N m² C−1
(b) The three faces that touch the corner all lie in planes passing through the charge. The field lines from the charge run along those planes, so E is parallel to each of these faces and perpendicular to their normals. Flux through each = zero.
(c) All the flux therefore passes through the remaining three faces, which are equivalent by symmetry:
φface = (1.13 × 105) ÷ 3 = 3.76 × 104 N m² C−1 through each.
Show Answer
• The two flat ends have normals along the wire, perpendicular to E, so they contribute zero flux.
• On the curved surface, E is parallel to the outward normal and constant in magnitude, so its flux is E(2πrl).
Total flux: φ = E(2πrl)
Charge enclosed: qenc = λl
Gauss’s theorem: E(2πrl) = λl ÷ ε0. The l cancels, giving
E = λ ÷ (2πε0r) = 2kλ ÷ r, directed radially outward for positive λ.
Numerical. 2kλ = 2(8.99 × 109)(2.0 × 10−6) = 3.596 × 104
At r = 0.10 m: E = (3.596 × 104) ÷ 0.10 = 3.60 × 105 N C−1
For E = 1.0 × 105 N C−1: r = 2kλ ÷ E = (3.596 × 104) ÷ (1.0 × 105) = 0.36 m = 36 cm
Show Answer
• The curved side has its normal lying in the plane of the sheet, perpendicular to E: flux = 0.
• Each flat end has its normal parallel to E: flux = EA each, giving 2EA in total.
Charge enclosed: qenc = σA
Gauss’s theorem: 2EA = σA ÷ ε0. The A cancels:
E = σ ÷ (2ε0), directed away from the sheet for positive σ.
(a) E = (1.2 × 10−6) ÷ [2(8.854 × 10−12)] = (1.2 × 10−6) ÷ (1.7708 × 10−11) = 6.78 × 104 N C−1
(b) It does not change. No r appears in the result — the field of an infinite sheet is uniform, the same at every distance.
(c) If two field lines intersected, two tangents could be drawn at the point of intersection, meaning the electric field would have two different directions at the same point. That is impossible, so field lines never cross.
Show Answer
φ = E × 4πr²
(i) Outside, r > R. The whole charge is enclosed: qenc = Q.
E × 4πr² = Q ÷ ε0 → E = Q ÷ (4πε0r²) = kQ ÷ r²
The shell behaves exactly like a point charge Q placed at its centre.
(ii) Inside, r < R. All the charge lies on the surface at radius R, which is outside the Gaussian sphere, so qenc = 0.
E × 4πr² = 0 → E = 0 everywhere inside the shell.
Numerical. kQ = (8.99 × 109)(8.0 × 10−6) = 7.192 × 104
(a) r = 0.30 m (outside): r² = 0.090 m²
E = (7.192 × 104) ÷ 0.090 = 7.99 × 105 N C−1
(b) r = 0.15 m (on the surface): r² = 0.0225 m²
E = (7.192 × 104) ÷ 0.0225 = 3.20 × 106 N C−1 — the maximum value.
(c) r = 0.10 m (inside): E = 0
Graph. E stays flat at zero from r = 0 up to r = R; at r = R it jumps to kQ ÷ R² = 3.20 × 106 N C−1; beyond that it falls off smoothly as 1 ÷ r², so doubling r from 0.15 m to 0.30 m divides the field by 4 (3.20 × 106 ÷ 4 = 8.0 × 105, matching part (a)). ✓
One More Than Yesterday
If some of that felt hard, good — that is what learning feels like from the inside. You do not need to master this chapter today. You need to be slightly better at it than you were yesterday. Get one more question right tomorrow than you did today, write the shell derivation one more time this week than last, and by the time the board exam arrives those small daily gains will have quietly compounded into real fluency. Small and steady beats heroic and occasional, every single time. See you in the next chapter — and if you want to switch subjects for a while, the rest of the Class 12 notes library is waiting.