State Gauss's law and use it to derive the electric field due to an infinitely long, uniformly charged wire.
From NCERTIn depthAsked by Cupe 1. from New Delhi, India23 Oct 2026, 11:21 PM ISTUpdated
In one line
Gauss's law helps us find the electric field around objects by looking at the total charge inside an imaginary surface.
Carl Friedrich Gauss — German polymath and scholar (1777–1855) Carl Friedrich Gauss, who formulated Gauss's law in 1835. Image: Wikipedia
The rule
Gauss's Law states that the electric flux through a closed surface is directly proportional to the charge enclosed within that surface.
Formula: ∮ E · dA = Q_enc / ε₀ Where:
∮ means sum over a closed surface
E is the electric field
dA is a tiny area on the surface
Q_enc is the total charge inside the surface
ε₀ (epsilon naught) is the permittivity of free space (a constant)
The formula
E = λ / (2πε₀r)
Eelectric field (N/C)
λcharge per unit length (C/m)
ε₀permittivity of free space (8.85 × 10⁻¹² C²/N·m²)
rdistance from the wire (m)
Let's understand it simply
Imagine you are standing in the middle of a cricket field, holding a big balloon filled with charge. You cover yourself with a giant, invisible plastic ball (the closed surface). Gauss's law says: if you count how much electric field passes through the plastic ball, it depends only on how much charge is inside, not outside.
For an infinitely long, charged wire, think of a long rope with charge spread evenly along it. To find the electric field near it, we use a tube (cylinder) as our imaginary surface, wrapping around the wire like a pipe. This helps us count the field lines coming out.
How it works
Choose a Gaussian surface: Pick a cylinder around the wire, with the wire running along its length. The cylinder has radius r and length L.
Find the charge enclosed: If the wire has charge per unit length λ (lambda), the total charge inside the cylinder is Q_enc = λ × L.
Calculate electric flux: The electric field points straight out from the wire. Only the curved side of the cylinder gets field lines; the flat ends get none (field is parallel to them).
Write Gauss's law: The total flux through the curved surface is E × (2πrL) (area of the curved side).
Set flux equal to charge/ε₀: So, E × (2πrL) = λL / ε₀.
Solve for E: Cancel L on both sides to get E = λ / (2πε₀r).
Finding Electric Field using Gauss's Law for a Wire
Pick a cylinder around the wire as Gaussian surface
Find charge inside: Q_enc = λ × L
Electric field points outwards, only through curved surface
Total flux: E × (2πrL)
Apply Gauss's law: E × (2πrL) = λL / ε₀
Solve: E = λ / (2πε₀r)
The story behind it
Gauss's law is named after Carl Friedrich Gauss, a German mathematician, who published it in 1835. He wanted to find a simple way to connect electric fields to the charges that make them. Gauss was famous for making hard problems simple, and this law is a great example. It is one of Maxwell's four equations, which are the foundation of all electricity and magnetism.
Where you see it in real life
High-voltage power lines: The field around the wire can be found using this law.
Charging a metal rod: The field near the rod is like the wire in this example.
Lightning rods: The field around the rod helps decide where lightning will strike.
Electric fences: The field around the wire keeps animals away.
Cables in electronics: Engineers use this law to design safe insulation.
Physics labs: Used to measure charge on long wires or rods.
Why it matters
Quick calculations: Lets you find electric fields without hard maths, if the object is symmetrical (like a wire, sphere, or plate).
Designing equipment: Used in making capacitors, cables, and insulators.
Understanding nature: Explains how charges spread out on conductors.
Exams: This derivation is a common question in board exams and entrance tests.
Worked example
Let's find the electric field 2 cm away from a wire with λ = 5 × 10⁻⁶ C/m.
Given: λ = 5 × 10⁻⁶ C/m, r = 2 cm = 0.02 m, ε₀ = 8.85 × 10⁻¹² C²/(N·m²)
Step 1: Write the formula E = λ / (2πε₀r)
Step 2: Plug in the numbers E = (5 × 10⁻⁶) / (2 × π × 8.85 × 10⁻¹² × 0.02)
Step 3: Calculate denominator 2 × π × 8.85 × 10⁻¹² × 0.02 ≈ 1.113 × 10⁻¹²
Step 4: Divide E = (5 × 10⁻⁶) / (1.113 × 10⁻¹²) ≈ 4.49 × 10⁶ N/C
Common mistakes
Wrong surface: Using a sphere or box instead of a cylinder for a wire. Always match the surface to the symmetry of the object.
Forgetting the ends: Including the flat ends of the cylinder in the flux. The field is parallel to these, so flux is zero there.
Missing units: Not converting cm to m, or forgetting units for charge and distance.
Mixing up λ and Q: λ is charge per length, not total charge.
Remember it
For an infinite wire, electric field drops as you move away: E = λ / (2πε₀r). Think: "More charge, more field; farther away, less field."
Check yourself
Why do we use a cylinder (not a sphere) as the Gaussian surface for a long wire?
Because the wire is long and straight, and the field spreads out evenly in circles around it.
If you double the distance from the wire, what happens to the electric field?
It becomes half, because E is inversely proportional to r.
What is λ in the formula E = λ / (2πε₀r)?
λ is the charge per unit length (C/m) on the wire.
Why is there no flux through the flat ends of the cylinder?
Because the electric field is parallel to those ends, so no field lines pass through them.
Quick recap
Gauss's law links electric field to charge inside a closed surface.
For an infinite wire, use a cylinder as the Gaussian surface.
Electric field from the wire: E = λ / (2πε₀r).
Only the curved side of the cylinder gets field lines.
Always match the surface to the object's symmetry.