Chapter 1 · Sections 1.13–1.14Textbook pages 32–36

Continuous Charge Distribution & Gauss's Law

Real charged objects – a charged rod, a metal sheet, a thundercloud – hold billions of billions of electrons. Instead of counting them one by one, we describe the charge by a smooth density. Then we meet one of the most powerful results in electrostatics: Gauss's law – the total flux out of any closed surface depends only on the charge inside it.

📖 What you need (pages 2, 9 and 10 of this guide)

Electric field (വൈദ്യുത മണ്ഡലം) of a point charge: E = 14πε₀ qr² r̂, directed away from a positive charge (page 9).
Flux through a small area: Δφ = E·ΔS = E ΔS cos θ; the area vector (ഏരിയ സദിശം) ΔS points along the outward normal of a closed surface (page 10).
Areas: circle πr², sphere 4πr², curved surface of a cylinder 2πrl (page 2). ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻².

1. Continuous charge distribution (1.13, Fig 1.24)

So far we have worked with a few discrete point charges (ബിന്ദു ചാർജ്) q₁, q₂, …, q_n. On the surface of a charged conductor, however, it is impractical to locate every electron. We treat the charge as a continuous charge distribution (തുടർച്ചയായ ചാർജ് വിതരണം): pick a small element that is tiny on the everyday (macroscopic) scale but still contains a very large number of electrons, and state how much charge ΔQ it holds.

Charge spread along…	Density	Definition	SI unit	Example
a line (thin wire, rod)	linear charge density λ രേഖീയ ചാർജ് സാന്ദ്രത	λ = ΔQΔl (1.24)	C/m (C m⁻¹)	charged thin wire
a surface (sheet, conductor surface)	surface charge density σ ഉപരിതല ചാർജ് സാന്ദ്രത	σ = ΔQΔS (1.23)	C/m²	charged metal plate, spherical shell
a volume (cloud, solid ball)	volume charge density ρ വ്യാപ്ത ചാർജ് സാന്ദ്രത	ρ = ΔQΔV (1.25)	C/m³	charged cloud, electron cloud of an atom

So the charge in an element is ΔQ = λΔl, ΔQ = σΔS or ΔQ = ρΔV. These densities are smoothed-out averages: they ignore the quantisation of charge and the empty space between the charges at the atomic level – just as we speak of the density of water and forget its molecules.

🧪 Virtual lab: zoom into a charge element (Fig 1.24)

element size mm

Change the element size: ΔQ changes, but ΔQ ÷ (size) – the density – stays the same for a uniform distribution. Even a 0.1 mm element contains lakhs of elementary charges, so a smooth density is a very good description.

Field of a continuous distribution

Divide the charge into small volume elements ΔV. The element at position r holds charge ρΔV and acts like a point charge. By Coulomb's law its field at the point P (position vector R) is

ΔE = 14πε₀ ρ ΔVr′² r̂′ (1.26)

where r′ is the distance from the element to P and r̂′ the unit vector from the element towards P. By the superposition principle (അധ്യാരോപണ തത്വം) the total field is the vector sum over all elements:

E ≅ 14πε₀ ∑_{all ΔV} ρ ΔVr′² r̂′ (1.27)

When ΔV → 0 the sum becomes an integral, E = 14πε₀ ∫ ρ dVr′² r̂′. Doing this sum directly is usually very hard – which is exactly why Gauss's law, below, is so useful.

🌩️ Real life – charge on a thundercloud

Nobody can count the electrons in a thundercloud. Meteorologists describe the cloud by a volume charge density ρ (a few nC per cubic metre in the charged regions) and the ground below by a surface charge density σ. The lightning conductor on a tall building works because charge on a sharp metal tip has a large σ and therefore a strong field near it (more on page 17).

2. Gauss's law – flux through a sphere around a point charge (1.14, Fig 1.25)

We now use the idea of electric flux (വൈദ്യുത ഫ്ലക്സ്) from page 10. Put a point charge q at the centre of a sphere of radius r, divide the sphere into small area elements ΔS and add up the flux through all of them. Step through the derivation – this is exactly how you should write it in the exam.

📝 Board exam: "State Gauss's law. Show that the flux through a sphere enclosing a point charge q is q/ε₀" – write these steps with the labelled diagram.

🎬 Derivation in exam layout – flux through a sphere enclosing q (Fig 1.25)

⭐ Gauss's law (ഗോസ് നിയമം) – Eq. 1.31

The total electric flux through any closed surface is equal to 1ε₀ times the total charge enclosed by that surface:

φ = ∮ E·dS = q_enclosedε₀

The closed surface used to apply the law is called a Gaussian surface (ഗോസിയൻ ഉപരിതലം). The ∮ sign means "add E·dS over the whole closed surface". If no charge is enclosed, the net flux is zero.

🤔 Why does the radius r cancel?

The field falls as 1/r² while the area of the sphere grows as r². Their product stays the same, so every sphere around q – big or small – has the same flux q/ε₀. In the picture of field lines (വൈദ്യുത മണ്ഡലരേഖകൾ): the same number of lines leaves q and crosses every sphere around it. This works only because Coulomb's law is an exact inverse square law (വിപരീത വർഗ നിയമം).

No charge inside → zero net flux (Fig 1.26)

Take a closed cylinder with its axis parallel to a uniform electric field (ഏകീകൃത വൈദ്യുത മണ്ഡലം). There is no charge inside, so Gauss's law predicts zero total flux. Check it face by face:

🎬 Zero flux through a cylinder in a uniform field (Fig 1.26)

Whenever the net flux through a closed surface is zero, the total charge inside it is zero (there may be equal + and − charges inside, but they add to zero).

3. Points to note about Gauss's law (remarks i–vi)

i Gauss's law is true for any closed surface, whatever its shape or size.

ii q on the right side is the sum of all charges enclosed by the surface (with signs). They may be anywhere inside.

iii If some charges are inside and some outside, the field E on the left side is due to all the charges, inside and outside. But q on the right side is only the charge inside.

iv The chosen surface is the Gaussian surface. It must not pass through a discrete charge (the field there is not defined – it grows without bound). It may pass through a continuous distribution.

v Gauss's law makes field calculations easy when the charges have symmetry – choose a Gaussian surface on which E is constant and along (or perpendicular to) the surface normal (page 13).

vi Gauss's law rests on the inverse square law in Coulomb's law. Any violation of Gauss's law would mean a departure from 1/r².

⚠ Common mistake

"Only the charges inside produce the field on the Gaussian surface" – wrong. Outside charges do change E at each point of the surface; they only add up to zero net flux, because every field line of an outside charge that enters the surface also leaves it. Try it in the lab below.

4. Gaussian-surface lab – drag charges in and out

🧪 Virtual lab: what does the flux meter count?

surface size

Drag the four charges (+2q, −1q, +1q, −2q). The small arrows show the resultant field E of all four charges – they change whenever you move any charge. The flux meter changes only when a charge crosses the surface. Turn on "own field lines" (8 lines per q): a line of an outside charge that enters the surface (red dot) always leaves it again (green dot). This is a cross-section through the centre; the charges lie in this plane.

5. Example 1.11 – flux through a cube (Fig 1.27)

Example 1.11. The electric field components in Fig 1.27 are E_x = αx^1/2, E_y = E_z = 0, in which α = 800 N C⁻¹ m^−1/2. Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that a = 0.1 m. (The cube of side a has its left face at x = a and its right face at x = 2a.)

E has only an x-component, so only the two faces perpendicular to x (left and right) have flux. On the left face the outward normal points along −x (θ = 180°); on the right face along +x (θ = 0°). The field is stronger on the right face, so more flux leaves than enters. Then use q = ε₀φ.

🎬 Solution in exam layout – Example 1.11

Answer: (a) φ = αa^5/2(√2 − 1) = 1.05 N m² C⁻¹ (outward). (b) q = ε₀φ = 9.27 × 10⁻¹² C (textbook value).
Note: carrying all the digits, φ = 1.048 N m² C⁻¹ and q = 8.854 × 10⁻¹² × 1.048 = 9.28 × 10⁻¹² C; using the rounded φ = 1.05 gives 9.30 × 10⁻¹² C. All of these round to ≈ 9.3 × 10⁻¹² C – any of them earns the mark.

6. Example 1.12 – flux through a cylinder (Fig 1.28)

Example 1.12. An electric field is uniform, and in the positive x direction for positive x, and uniform with the same magnitude but in the negative x direction for negative x. It is given that E = 200 î N/C for x > 0 and E = −200 î N/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x = +10 cm and the other is at x = −10 cm (Fig 1.28). (a) What is the net outward flux through each flat face? (b) What is the flux through the side of the cylinder? (c) What is the net outward flux through the cylinder? (d) What is the net charge inside the cylinder?

On both flat faces the field points outwards (to the left on the left face, to the right on the right face), so both fluxes are positive. Area of a face = πr² with r = 0.05 m. On the curved side, E ⟂ ΔS.

🎬 Solution in exam layout – Example 1.12

Answer: (a) +1.57 N m² C⁻¹ through each flat face; (b) zero through the side; (c) φ = 3.14 N m² C⁻¹; (d) q = ε₀φ = 2.78 × 10⁻¹¹ C (positive – such a field is produced by a positively charged sheet in the y-z plane, page 13).

🧲 Real life – Gauss's law in a laboratory

Physicists have tested Gauss's law (and so the inverse square law) by charging a closed metal shell and checking for any field inside it with sensitive meters. Such experiments show that the power of r in Coulomb's 1/r² law equals 2 to within about one part in 10¹⁶ – remark (vi) in action. The same idea keeps a car safe in lightning (a Faraday cage – page 17).

🌐 Explore online needs internet · opens in a new tab

📄 Electric flux Wikipedia
Flux of a field through a surface.
📄 Gauss's law Wikipedia
Gauss's law and its applications to symmetric charge distributions.
📄 Carl Friedrich Gauss Wikipedia
The "prince of mathematicians" behind Gauss's law.
📄 Faraday cage Wikipedia
Why the field inside a closed conductor is zero – electrostatic shielding (Exercise 1.28).

7. Check yourself

🧭 Summary

Continuous distributions: λ = ΔQ/Δl (C/m), σ = ΔQ/ΔS (C/m²), ρ = ΔQ/ΔV (C/m³). Elements are small macroscopically but hold huge numbers of charges.
Field of a distribution: E = (1/4πε₀) ∑ ρΔV r̂′/r′² (superposition).
Sphere around a point charge: φ = q/(4πε₀r²) × 4πr² = q/ε₀, independent of r.
Gauss's law: φ = ∮E·dS = q_enclosed/ε₀ for any closed surface. Outside charges give zero net flux, but E is due to all charges.
Ex 1.11: φ = 1.05 N m² C⁻¹, q = 9.27 × 10⁻¹² C. Ex 1.12: 1.57 + 1.57 + 0 = 3.14 N m² C⁻¹, q = 2.78 × 10⁻¹¹ C.

Next: Applications of Gauss's law – wire, sheet and spherical shell ▶