A finite line of charge · Anannt Study

Integrating the field of a line charge

You will be able to: Write $dq=\lambda dx$, resolve $d\vec{E}$, and integrate a finite straight line; recover the infinite-line limit.

1. Model

Name the physical situation, the idealisations, and the quantities that matter.

2. Represent

Draw the diagram, choose coordinates, and write the symbols you will use.

3. Derive

Select a principle and produce a result from it, keeping the chain of reasoning visible.

4. Check

Units, signs, limiting cases, and whether the result could be true of the situation.

5. Explain

Say what the result means, what it assumed, and why a tempting alternative fails.

Reviewed 2026-09-10 · v1.0 · Anannt physics author (author) · Independent physics reviewer (reviewer) · Video is never required for mastery.

2. Explanation (text route)

A rod is divided into dx. At P on the bisector, paired dE arrows show cancellation along the rod and addition along the bisector. The infinite-line 1/r result is compared with a graph of E(r).

A uniform line is an idealised cylinder of radius 0 with finite $λ$ . Real wires have radius; we stay many radii away. The infinite line ignores ends.

3. Worked example

Finite rod, perpendicular bisector

Situation. Rod length $2 L = 0.40 m$ , $λ = 3.0 μ C/m$ , point $P$ a distance $d = 0.15 m$ from the centre on the bisector. Find $E$ .

Assumptions. Uniform $\lambda$; vacuum; rod thickness neglected.

Derivation. $E = \frac{1}{4 π ε _{0}} \frac{λ ( 2 L )}{d d ^{2} + L ^{2}} = k λ (2 L) / (d d^{2} + L^{2})$ . Numerically $k = 8.99 e 9$ , $2 L = 0.40$ , $0.1 5^{2} + 0.2 0^{2} = 0.250$ , $E = (8.99 e 9) (3.0 e - 6) (0.40) / (0.15 \times 0.250) = 2.88 \times 1 0^{5} N/C$ , away from the rod if $λ > 0$ .

Units. N/C.

Interpretation. Compare with a point charge $Q = λ (2 L)$ at the centre: $k Q / d^{2} = 4.79 \times 1 0^{5} N/C$ , larger, as expected — the actual charge is spread farther away.

Why tempting alternatives fail. Infinite-line formula $λ / (2 π ε_{0} d) = 3.6 \times 1 0^{5} N/C$ is a different idealisation (no ends). Using it here is an approximation, not the finite-rod result.

4. Faded example

Same rod, field point on the axis beyond one end.

Do perpendicular components cancel?

Hint: There is no left-right pairing about P.

On the axis of a straight rod, $d E$ is along the axis; there is no perpendicular component to cancel. You integrate a 1-D scalar.

What is $E$ as $P$ approaches an end?

Complete the previous step first.

6. Independent questions (different situations)

These submissions are required for evidence. Watching or reading the explanation does not advance mastery.

An infinite line has $λ = 2.0 μ C/m$ . $E$ at $r = 0.10 m$ is closest to

3.6 \times 1 0^{5} N/C

1.8 \times 1 0^{5} N/C

1.8 \times 1 0^{4} N/C

k λ / r^{2}

If the perpendicular distance from an infinite line charge doubles, $E$ is multiplied by

A.1/4B.1/2C.2D.unchanged, like a sheet