An infinite line charge produces a field of $9 \times 10^{4}\,\text{N/C}$ at a distance of $2\,\text{cm}$. Calculate the linear charge density.

An infinite line charge produces a field of 9 × 10⁴N/C at a distance of 2cm. Calculate the linear charge densi

Exercises · 1.22

9 \times 10^{4}\,\text{N/C}

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Concept understanding — Gauss Law

Gauss's Law is a shortcut. Instead of adding up the Coulomb contribution of every charge — a nightmare of vectors and integrals — it lets you read the field straight off the symmetry of a problem. The whole idea rests on one quantity: electric flux.

Flux — field-lines counted through a surface. For a uniform field E crossing a flat area A, the flux is Φ = E·A = EA cosθ, where θ is the angle between the field and the normal to the surface. Picture the field as a bundle of lines; flux is how many pierce the surface. When E is edge-on (θ = 90°) nothing passes through and Φ = 0; when E is ⊥ to the surface (θ = 0) the count is maximal. For a closed surface, Φ is the net number of lines leaving it — lines that enter and exit cancel.

The law itself. Gauss's law states that the net flux out of any closed surface depends only on the charge trapped inside:

Φ = q_enclosed / ε₀.

Two consequences do most of the work. First, only enclosed charge counts — a charge outside sends as many lines in as out, so its net contribution is exactly zero. Second, the surface's shape is irrelevant; move the charge around inside or deform the surface, and Φ never changes.

Why symmetry makes it powerful. By itself Φ = q/ε₀ has E buried in an integral. It becomes a tool only when you pick a Gaussian surface matched to the symmetry — one where E is constant and everywhere either ⊥ to the surface (so Φ = EA) or ∥ to it (contributing nothing). Then E slides out and you solve in one line. This works for exactly three geometries:

1 — Infinite line charge (linear density λ). Use a coaxial cylinder: E = λ / 2πε₀r, falling off as ∝ 1/r.

2 — Infinite sheet (surface density σ). Use a pillbox pierced through the sheet: E = σ / 2ε₀ — uniform and completely independent of distance. The field near a large charged plane simply doesn't weaken as you step back. …

Why this formula?

Gauss's Law: Why It Holds

Gauss's Law is one of the four Maxwell's equations and a cornerstone of electromagnetism. Let's build the understanding from the ground up — not just the formula, but the why.

1. The Core Idea: Flux as "Flow" of Field

Imagine an electric field $\vec{E}$ passing through a small patch of area $dA$ . The electric flux through that patch is:

d\Phi_E = \vec{E} \cdot d\vec{A} = E \, dA \cos\theta

where $\theta$ is the angle between $\vec{E}$ and the outward normal to the surface.

Why this definition?

If $\vec{E}$ is perpendicular to the surface ( $\theta = 0$ ), maximum field "flows through".
If $\vec{E}$ is parallel ( $\theta = 90^\circ$ ), no flux — the field just slides along the surface.

Total flux through a closed surface $S$ is:

\Phi_E = \oint_S \vec{E} \cdot d\vec{A}

2. The Key Insight: Flux Depends Only on Enclosed Charge

Consider a single point charge $+q$ at the centre of a spherical surface of radius $r$ .

By Coulomb's law, at every point on the sphere: $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$ , radially outward.
The area vector $d\vec{A}$ is also radially outward.
So $\vec{E} \cdot d\vec{A} = E \, dA$ everywhere.

The total flux:

\Phi_E = \oint E \, dA = E \oint dA = \left( \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \right) (4\pi r^2) = \frac{q}{\varepsilon_0}

Notice: The $r^2$ cancels! The flux is independent of the radius.

3. Why Shape Doesn't Matter

Now, what if the surface is not spherical but any closed shape enclosing the charge?

Draw a small cone from the charge to the surface.
The flux through a patch $dA$ at distance $r$ is $d\Phi = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \cos\theta \, dA$ .
But $\frac{\cos\theta \, dA}{r^2}$ is exactly the solid angle $d\Omega$ subtended by $dA$ at the charge.

So $d\Phi = \frac{q}{4\pi\varepsilon_0} d\Omega$ .

Integrating over the entire closed surface: $\oint d\Omega = 4\pi$ (total solid angle around a point).

Hence:

\Phi_E = \frac{q}{4\pi\varepsilon_0} \cdot 4\pi = \frac{q}{\varepsilon_0}

Result: For any closed surface enclosing $q$ , the flux is $\frac{q}{\varepsilon_0}$ .

4. Multiple Charges: Superposition

If there are many charges $q_1, q_2, \dots, q_n$ inside the surface, the total electric field is the vector sum of individual fields:

\vec{E} = \vec{E}_1 + \vec{E}_2 + \dots + \vec{E}_n

Flux is linear: …

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Applying Gauss's law to a coaxial cylinder gives $E=\lambda/2\pi\varepsilon_0 r$ ; inverting for the given $E$ and $r$ yields $\lambda=2\pi\varepsilon_0 rE=1.0\times10^{-7}\,\text{C/m}=0.1\,\mu\text{C/m}$ .

An infinite line charge has cylindrical symmetry: the field is radial and its magnitude depends only on the perpendicular distance $r$ . This makes a coaxial cylinder the natural Gaussian surface.

Step 1 — Gaussian surface. Take a cylinder of radius $r$ and length $L$ coaxial with the line. It encloses charge $q_{\text{enc}}=\lambda L$ .

Step 2 — Flux. $\vec E$ is perpendicular to the curved wall (area $2\pi r L$ ) and parallel to the flat end caps (zero flux there), so

\Phi_E=E(2\pi r L).

Step 3 — Gauss's law.

E(2\pi r L)=\frac{\lambda L}{\varepsilon_0}\ \Rightarrow\ E=\frac{\lambda}{2\pi\varepsilon_0 r},

the length $L$ cancelling, as it must for an infinite line. …

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Common Mistakes with Gauss Law: Line Charge Problem

Mistake 1: Using the Wrong Formula for Field of a Line Charge

The error: Students often confuse the electric field formulas for different charge distributions — using the point charge formula ( $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$ ) or the infinite sheet formula ( $E = \frac{\sigma}{2\varepsilon_0}$ ) instead of the line charge formula.

Why it's wrong: Each charge distribution has a unique symmetry, and Gauss Law gives a different result for each. For an infinite line charge, the field falls off as $1/r$ , not $1/r^2$ .

How to avoid: Memorise the three standard results from Gauss Law:

Point charge: $E \propto \frac{1}{r^2}$
Infinite line charge: $E \propto \frac{1}{r}$
Infinite sheet: $E$ is constant (independent of distance)

For this problem, the correct formula is:

E = \frac{\lambda}{2\pi\varepsilon_0 r}

where $\lambda$ is the linear charge density.

Mistake 2: Forgetting to Convert Units

The error: Using $r = 2$ directly in the formula without converting cm to m.

Why it's wrong: All SI units must be consistent. The electric field is in N/C, $\varepsilon_0$ has units of $\text{C}^2/\text{N·m}^2$ , so distance must be in metres.

How to avoid: Always write the conversion step explicitly:

r = 2\,\text{cm} = 2 \times 10^{-2}\,\text{m} = 0.02\,\text{m}

Mistake 3: Incorrect Value or Units of $\varepsilon_0$

The error: Using $\varepsilon_0 = 8.85 \times 10^{-12}$ but forgetting the units, or using $9 \times 10^9$ (which is $\frac{1}{4\pi\varepsilon_0}$ ) without adjusting the formula.

Why it's wrong: The formula $E = \frac{\lambda}{2\pi\varepsilon_0 r}$ uses $\varepsilon_0$ directly. If you use $k = \frac{1}{4\pi\varepsilon_0}$ , the formula becomes:

E = \frac{2k\lambda}{r}

Both are correct — but mixing them up gives wrong answers.

How to avoid: Stick to one consistent form throughout the calculation. I recommend:

E = \frac{\lambda}{2\pi\varepsilon_0 r}

with $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N·m}^2$ .

Mistake 4: Algebraic Errors While Solving for $\lambda$

The error: Rearranging incorrectly — for example, writing $\lambda = E \times 2\pi\varepsilon_0 r$ instead of $\lambda = E \times 2\pi\varepsilon_0 r$ (this one is actually correct, but students often multiply/divide the wrong terms).

How to avoid: Write the rearrangement step-by-step:

$E = \frac{\lambda}{2\pi\varepsilon_0 r}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AHSEC Higher Secondary (HS) Final Examination 2024Set ANNUAL2 marks
$Q. Classify the following into$
(i) polar, and
(ii) non-polar molecules: CO2, HCl, H2 and H2O. OR Calculate the flux passing through a circular area of radius 5cm placed perpendicular to a uniform electric field E = 200i NC^-1.
›Reveal solutionSolution
Option 1: HCl and H₂O are polar (asymmetric charge distribution, net dipole moment); CO₂ and H₂ are non-polar (symmetric, zero net dipole moment). Option 2: electric flux $\Phi = EA \approx 1.57\,\mathrm{N\,m^2\,C^{-1}}$ .

Option 1 — Polar vs non-polar molecules

A molecule is polar if its centres of positive and negative charge do not coincide, giving it a permanent (net) dipole moment; it is non-polar if they coincide (net dipole moment zero), usually due to symmetry.

CO₂ — linear, symmetric ( $O=C=O$ ); the two C=O bond dipoles are equal and opposite, so they cancel $\Rightarrow$ non-polar.

HCl — different atoms (H, Cl) with unequal electronegativity, bond dipole does not cancel $\Rightarrow$ polar.

H₂ — homonuclear diatomic (identical atoms), no charge asymmetry $\Rightarrow$ non-polar.

H₂O — bent (angular) shape; the two O–H bond dipoles do not cancel $\Rightarrow$ polar.

So: Polar: HCl, H₂O. Non-polar: CO₂, H₂.

Option 2 — Electric flux

…
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AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL2 marks
$Q. A closed spherical surface encloses a charge q at its centre. Show that electric flux through the closed surface is q/ε0. OR A pair of charges +q and -q, separated by a small distance 2a is placed in an electric field E, so that the line joining the charges makes an angle θ with E. Write the expressions for torque τ and also its magnitude |τ|.$
›Reveal solutionSolution

Gauss's law shows the flux out of a sphere enclosing a point charge q is q/ε₀, independent of the sphere's radius; and the torque on a dipole in a field is τ = pE sinθ.

Main question: Consider a point charge $q$ at the centre O of a spherical surface of radius $r$ . By Coulomb's law, the magnitude of the electric field at every point on this sphere (all at distance $r$ from $q$ ) is the same:

$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$

and it points radially outward, i.e. parallel to the outward area vector $d\vec A$ at every point on the sphere (by symmetry). So the flux through the whole closed surface is simply $E$ times the total surface area $4\pi r^2$ :

$\Phi_E = \oint \vec E \cdot d\vec A = E \times 4\pi r^2 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\times 4\pi r^2 = \frac{q}{\epsilon_0}$

Notice the radius $r$ cancels out completely — the flux depends only on the enclosed charge $q$ , not on the size of the sphere. This is exactly Gauss's law for this special symmetric case.

…

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Q.An infinite line charge produces a field of $9 \times 10^{4}\,\text{N/C}$ at a distance of $2\,\text{cm}$ . Calculate the linear charge density.

Concept understanding — Gauss Law

Why this formula?

Gauss's Law: Why It Holds

1. The Core Idea: Flux as "Flow" of Field

2. The Key Insight: Flux Depends Only on Enclosed Charge

3. Why Shape Doesn't Matter

4. Multiple Charges: Superposition

Method: Gauss's Law for an Infinite Line Charge

Why Gauss's Law?

Steps

Numerical Calculation

Common Mistakes with Gauss Law: Line Charge Problem

Mistake 1: Using the Wrong Formula for Field of a Line Charge

Mistake 2: Forgetting to Convert Units

Mistake 3: Incorrect Value or Units of $\varepsilon_0$

Mistake 4: Algebraic Errors While Solving for $\lambda$

Q.An infinite line charge produces a field of 9×104 N/C9 \times 10^{4}\,\text{N/C}9×104N/C at a distance of 2 cm2\,\text{cm}2cm. Calculate the linear charge density.

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Q.An infinite line charge produces a field of $9 \times 10^{4}\,\text{N/C}$ at a distance of $2\,\text{cm}$ . Calculate the linear charge density.

Unlock everything free for 14 days