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.

For an infinite line charge, Gauss's law with a coaxial cylinder gives

With $E=9\times10^{4}\,\text{N/C}$, $r=2\,\text{cm}=0.02\,\text{m}$, and $\dfrac{1}{2\pi\varepsilon_0}=2\times9\times10^{9}$: …

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

Step 3 — Gauss's law.

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

Method: Gauss's Law for an Infinite Line Charge

Why Gauss's Law?

For an infinite line charge, the electric field has cylindrical symmetry — it points radially outward and depends only on the perpendicular distance from the wire. This symmetry makes a cylindrical Gaussian surface the perfect choice.

Steps

Step 1: Choose a Gaussian surface

Take a right circular cylinder of radius $r$ and length $L$, coaxial with the line charge.

Step 2: Apply Gauss's Law

Gauss's Law states:

• The field $\vec{E}$ is radial and constant in magnitude over the curved surface.
• Flux through the flat ends is zero (field is parallel to the surface).
• Only the curved surface contributes.

Step 3: Compute the flux

Area of curved surface = $2\pi r L$

Flux = $E \cdot (2\pi r L)$

Step 4: Find enclosed charge

If $\lambda$ is the linear charge density (charge per unit length), then:

Step 5: Equate and solve for $\lambda$

Cancel $L$:

So:

Numerical Calculation

Given:

• $E = 9 \times 10^{4}\,\text{N/C}$
• $r = 2\,\text{cm} = 0.02\,\text{m}$ …

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:

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:

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:

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

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

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:

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