Applications of Gauss's Law
For most charge distributions the sum (integral) of Coulomb fields is too hard to do. But when the charges have symmetry, Gauss's law gives the field in three or four lines. Here you derive the three classic results – the long charged wire, the infinite charged sheet and the charged spherical shell – and use them on Example 1.13, an early model of the atom.
- Gauss's law (ഗോസ് നിയമം): φ = ∮E·dS = qenclosed/ε0 for any closed surface – the Gaussian surface (ഗോസിയൻ ഉപരിതലം).
- Flux through an area: E ΔS cos θ – maximum (EΔS) when E is along the normal, zero when E is parallel to the surface.
- Densities: λ (C/m), σ (C/m2), ρ (C/m3). Areas: curved surface of a cylinder 2πrl, sphere 4πr2; volume of a sphere (4/3)πr3.
1. The recipe – using symmetry
To find the electric field (വൈദ്യുത മണ്ഡലം) with Gauss's law:
- Use symmetry to find the direction of E and what its magnitude can depend on.
- Choose a Gaussian surface through the point P on which E is either along the normal with constant magnitude, or parallel to the surface (zero flux).
- Write the flux φ = E × (area where E is normal). Find qenclosed.
- Apply Gauss's law φ = qenclosed/ε0 and solve for E. Write it as a vector.
📝 Board exam: each of the three derivations below is a favourite 5-mark long-answer question (Part D). Draw the labelled Gaussian surface – it carries marks.
2. Field due to an infinitely long straight uniformly charged wire (1.15.1, Fig 1.29)
Consider an infinitely long straight wire (അനന്തമായ നീളമുള്ള നേർകമ്പി) with uniform linear charge density (രേഖീയ ചാർജ് സാന്ദ്രത) λ. Step through the derivation:
E = λ2πε0r n̂ i.e. E ∝ 1r
n̂ is the radial unit vector (യൂണിറ്റ് സദിശം) in the plane normal to the wire through the point. E points outward if λ > 0 and inward if λ < 0.
- In E = E n̂ the scalar E is an algebraic number: if E < 0 (λ negative), E points opposite to n̂. The modulus |E| ≥ 0 is the magnitude.
- Only the charge λl inside the cylinder entered Gauss's law, but E is due to the charge on the entire wire.
- "Infinitely long" is essential – otherwise E would not be normal to the curved surface. For a real long wire the result holds near its middle, away from the ends.
3. Field due to a uniformly charged infinite plane sheet (1.15.2, Fig 1.30)
Let σ be the uniform surface charge density (ഉപരിതല ചാർജ് സാന്ദ്രത) of an infinite plane sheet (അനന്തമായ സമതല ഷീറ്റ്) lying in the y-z plane, so the x-axis is normal to it.
E = σ2ε0 n̂ – independent of the distance from the sheet
n̂ is the unit vector normal to the sheet, pointing away from it (on both sides). E points away from the sheet if σ > 0 and towards it if σ < 0. For a large but finite sheet this holds in the middle region, away from the edges.
An infinite uniform sheet of charge in the y-z plane looks exactly the same if you (a) translate it parallel to the y-z plane in any direction, or (b) rotate it about the x-axis through any angle. Since the system is unchanged, its field must be unchanged too:
- Translation along y ⇒ E at (0, y1, 0) equals E at (0, y2, 0). Translation along z ⇒ the same for (0, 0, z1) and (0, 0, z2).
- Rotation about the x-axis ⇒ E cannot have any component in the y-z plane (that component would rotate), so E is parallel to x.
- Gauss's law then adds that the magnitude does not depend on x either: the field is the same at all points, pointing in opposite directions on the two sides of the sheet. Compare this with the effort of a direct Coulomb's-law calculation!
4. Field due to a uniformly charged thin spherical shell (1.15.3, Fig 1.31)
A thin spherical shell (നേർത്ത ഗോളീയ കവചം) of radius R carries a uniform surface charge density σ, so its total charge is q = 4πR2σ. The situation has spherical symmetry: at any point P, E can depend only on r (the distance from the centre O) and must be radial.
Outside (r > R): E = q4πε0r2 r̂ Inside (r < R): E = 0
Outside, the shell acts as if all its charge were a point charge (ബിന്ദു ചാർജ്) at its centre. Inside, the field is zero everywhere. Just outside the surface E = q/(4πε0R2) = σ/ε0. The experimental check that E = 0 inside confirms the 1/r2 dependence in Coulomb's law (compare the uniform mass shell in gravitation, Class XI).
Charge on a closed metal body sits on its outer surface, and the field inside the enclosure is zero – the same idea as E = 0 inside a charged shell. That is electrostatic shielding (സ്ഥിതവൈദ്യുത കവചനം): a car with a metal body, an aircraft, or the metal-walled room around an MRI scanner protects what is inside. (Stay inside with the windows closed; do not touch metal parts. Details on page 17.)
5. Field explorer – E against distance for the three cases
Move the distance slider and watch the marker on the graph: the wire's field falls as 1/r, the sheet's stays constant, and the shell's is zero inside and falls as 1/r2 outside (ε0 = 8.854 × 10−12 C2 N−1 m−2).
| Charge distribution | Gaussian surface | Flux | qenclosed | Field E | Depends on distance as |
|---|---|---|---|---|---|
| Infinite line, λ | coaxial cylinder (r, l) | E × 2πrl | λl | λ/(2πε0r) | 1/r |
| Infinite plane sheet, σ | box (or cylinder) across the sheet, end area A | 2EA | σA | σ/(2ε0) | independent of distance |
| Thin spherical shell, q (outside) | concentric sphere, r > R | E × 4πr2 | q | q/(4πε0r2) | 1/r2 |
| Thin spherical shell (inside) | concentric sphere, r < R | E × 4πr2 | 0 | 0 | – |
- "The field of a sheet gets weaker further away" – no, for an infinite sheet it is σ/(2ε0) at every distance.
- "The field inside a charged shell is not zero" – it is zero: the Gaussian sphere inside encloses no charge.
- Writing 4πε0 instead of 2πε0 for the wire, or ε0 instead of 2ε0 for the sheet.
6. Example 1.13 – an early model of the atom (Fig 1.32)
Example 1.13. An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, what is the electric field at a distance r from the nucleus?
The negative charge is a uniform volume charge density (വ്യാപ്ത ചാർജ് സാന്ദ്രത) ρ with total −Ze, so (4/3)πR3ρ = −Ze. Use a spherical Gaussian surface centred on the nucleus. Inside (r < R) it encloses Ze plus only the part of the negative charge within radius r; outside (r > R) it encloses the whole neutral atom.
E is in units of Ze/(4πε0R2). Near the nucleus the atom's field is almost that of the bare nucleus (dashed); further out the enclosed negative charge cancels more and more of the nuclear charge, until at r = R it cancels completely.
- 📄 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
- Symmetry fixes the direction of E; a well-chosen Gaussian surface makes ∮E·dS = E × area.
- Infinite line: E = λ/(2πε0r) n̂ (∝ 1/r). Infinite plane sheet: E = σ/(2ε0) n̂ (independent of distance).
- Thin spherical shell: outside E = q/(4πε0r2) r̂ (as if all charge were at the centre); inside E = 0.
- Ex 1.13: E = (Ze/4πε0)(1/r2 − r/R3) for r < R; E = 0 for r > R.
- Symmetry operations (translation, rotation) that leave a system unchanged leave its field unchanged.