Electricity-And-Magnetism · Unit 4 · Video 1 · Interactive Practice

Charge Density and the Field of a Charged Rod (On Its Axis)

IKey Formulas

Formula	Name	What it means
$\rho = \dfrac{dq}{dV}$ , $\ \sigma = \dfrac{dq}{dA}$ , $\ \lambda = \dfrac{dq}{d\ell}$	Charge densities	Charge per volume ( $\mathrm{C/m^3}$ ), per area ( $\mathrm{C/m^2}$ ), per length ( $\mathrm{C/m}$ )
$Q = \int_V \rho\,dV$ ; uniform rod: $\lambda = \dfrac{Q}{\ell}$	Total charge	Integrate the density over the region; uniform density times size
$\vec{E} = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\int \frac{dq}{r^2}\,\hat{r}$	Field of a continuous charge	$r$ and $\hat{r}$ run from each element $dq$ to $P$ ; one integral per component
$\vec{E} = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{x_0(\ell + x_0)}\,\hat{i}$	Uniform rod, $P$ on its axis	$P$ at the origin, a distance $x_0$ from the near end; field points away from the rod

Key Insight: On the rod's axis every $d\vec{E}$ points along $-\hat{i}$ , so the vector integral collapses to one scalar integral, $\int_{x_0}^{x_0+\ell} dx'/x'^2$ .

IIVisualization 1 — A Vector Integral, Not a Sum of Magnitudes

Contributions add as vectors: $|\vec{E}| \le \int |d\vec{E}|$ , with equality only when every $d\vec{E}$ points the same way.

IIIVisualization 2 — Slicing the Rod: From Sum to Integral

Each slice is a point charge $\lambda\,dx'$ ; the sum of the slices' fields converges to $\dfrac{Q}{4\pi\varepsilon_0\,x_0(x_0+\ell)}$ .

Slices N 4

IVVisualization 3 — Far From the Rod: A Point Charge Again

As $x_0/\ell$ grows the exact field approaches a point charge $Q$ — fastest when that charge sits at the rod's centre.

Distance x₀/ℓ 10

💡 The centre wins because the rod's charge lies, on average, half a rod length beyond the near end: with $d = x_0 + \ell/2$ (so $x_0 = d - \ell/2$ and $x_0 + \ell = d + \ell/2$ ), the on-axis result rewrites as $x_0(x_0+\ell) = (d - \ell/2)(d + \ell/2) = d^2 - \ell^2/4$ , so the exact field is $\dfrac{Q}{4\pi\varepsilon_0\,(d^2 - \ell^2/4)}$ — slightly stronger than $\dfrac{Q}{4\pi\varepsilon_0 d^2}$ .

VQuiz Questions

Problem 1 · Linear Charge Density

Given: a thin rod of length $\ell = 0.50\ \text{m}$ carries a uniform total charge $Q = 4.0\ \text{nC}$ . Find its linear charge density $\lambda$ and the charge $dq$ on a $2.0\ \text{cm}$ piece of it.

What is $\lambda$ ?

What is $dq$ on the $2.0\ \text{cm}$ piece?

✅ Correct!

\lambda = Q/\ell = 8.0\ \text{nC/m}

, and a piece of length

dx'

holds

dq = \lambda\,dx' = (8.0\ \text{nC/m})(0.020\ \text{m}) = 0.16\ \text{nC}

❌ Not quite.

Q\ell = 2.0\ \text{nC}\cdot\text{m}

has the wrong units: a density is charge per length,

Q/\ell

❌ Not quite. That divides by the length twice (

Q/\ell^2 = 16\ \text{nC/m}^2

). A linear density divides by

\ell

once.

❌ Upside down.

\ell/Q = 0.125\ \text{m/nC}

is length per charge; the density is

Q/\ell

❌ Check $\lambda$ .

0.040\ \text{nC}

comes from using

\lambda = 2.0\ \text{nC/m}

; with

\lambda = 8.0\ \text{nC/m}

dq = \lambda\,dx'

is four times larger.

❌ Convert the length.

2.0\ \text{cm} = 0.020\ \text{m}

, not

2.0\ \text{m}

❌ That is the whole rod. The piece is

0.020/0.50 = 4\%

of the rod, so it holds

4\%

Q

❌ Not quite. Use

\lambda = Q/\ell

, then

dq = \lambda\,dx'

with

dx'

in metres.

Show solution

The charge is uniform, so the density is the total charge over the total length:

\lambda = \frac{Q}{\ell} = \frac{4.0\ \text{nC}}{0.50\ \text{m}} = 8.0\ \text{nC/m}

A piece of length $dx' = 2.0\ \text{cm} = 0.020\ \text{m}$ holds

dq = \lambda\,dx' = (8.0\ \text{nC/m})(0.020\ \text{m}) = 0.16\ \text{nC}

Check: $0.020\ \text{m}$ is $4\%$ of the rod, and $4\%$ of $4.0\ \text{nC}$ is $0.16\ \text{nC}$ .

Problem 2 · Field on the Axis

Given: a rod of length $\ell = 0.30\ \text{m}$ with uniform charge $Q = 5.0\ \text{nC}$ ; point $P$ lies on its axis, $x_0 = 0.20\ \text{m}$ from the near end. Take $\dfrac{1}{4\pi\varepsilon_0} = 8.99\times 10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$ . Find the field magnitude at $P$ .

✅ Correct!

E = \dfrac{Q}{4\pi\varepsilon_0\,x_0(\ell + x_0)} = \dfrac{44.95}{(0.20)(0.50)} \approx 450\ \text{N/C}

, pointing along the axis away from the rod.

❌ That treats the rod as a point charge at its near end.

Q/(4\pi\varepsilon_0 x_0^2)

puts all the charge

0.20\ \text{m}

away, but most of it is farther, so it overestimates. At

P

the rod is not far away:

x_0

is smaller than

\ell

❌ That is a point charge at the rod's centre.

Q/(4\pi\varepsilon_0 d^2)

with

d = 0.35\ \text{m}

ignores the

\ell^2/4

correction: with

x_0 = d - \ell/2

x_0(x_0+\ell) = d^2 - \ell^2/4

, so the exact field is

Q/\big(4\pi\varepsilon_0(d^2 - \ell^2/4)\big)

❌ That puts all the charge at the far end.

Q/\big(4\pi\varepsilon_0(x_0+\ell)^2\big)

underestimates. Each element sits at its own distance

x'

, and

\dfrac{1}{\ell}\displaystyle\int_{x_0}^{x_0+\ell}\frac{dx'}{x'^2} = \dfrac{1}{x_0(x_0+\ell)}

Show solution

From the integral over the rod, with $P$ at the origin and the rod from $x' = x_0$ to $x' = x_0 + \ell$ :

E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{\ell}\int_{x_0}^{x_0+\ell}\frac{dx'}{x'^2} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{x_0(\ell + x_0)}

Numerically, $\dfrac{Q}{4\pi\varepsilon_0} = (8.99\times10^{9})(5.0\times10^{-9}) = 44.95\ \text{N}\cdot\text{m}^2/\text{C}$ , so

E = \frac{44.95}{(0.20)(0.50)} = \frac{44.95}{0.10} \approx 450\ \text{N/C}

The charge is positive, so $\vec{E}$ points along the axis away from the rod: $\vec{E} \approx -450\,\hat{i}\ \text{N/C}$ .

Problem 3 · How Far Is "Far"?

Given: a uniform rod of length $\ell$ , charge $Q$ , and $P$ on its axis. Find the smallest distance at which the exact field differs by at most $1\%$ from the field of a point charge $Q$ placed (i) at the rod's near end, a distance $x_0$ from $P$ , and (ii) at the rod's centre, a distance $d$ from $P$ .

(i) Point charge at the near end: smallest $x_0$ ?

(ii) Point charge at the centre: smallest $d$ ?

✅ Correct! The near-end approximation needs

x_0 = 99\ell

, but the centre approximation is already within

1\%

d \approx 5.0\ell

— about twenty times closer.

❌ That is the $10\%$ threshold.

x_0/(x_0+\ell) = 0.9

gives

x_0 = 9\ell

; for

1\%

set the ratio to

0.99

❌ The ratio is not squared.

E_{\text{exact}}/E_{\text{point}} = \dfrac{x_0^2}{x_0(x_0+\ell)} = \dfrac{x_0}{x_0+\ell}

— one power, so

0.99

, not

\sqrt{0.99}

❌ At $x_0 = 10\ell$ the ratio is $10/11 \approx 0.91$ — still

9\%

off.

❌ Take the square root. The condition gives

d^2 = 25.25\,\ell^2

, so

d \approx 5.0\ell

❌ Check the correction term. The exact field has

d^2 - \ell^2/4

, not

d^2 - \ell^2

; with

\ell^2/4

the threshold is about half as far.

❌ The two approximations are not equally good. The centre version's error falls off like

\ell^2/(4d^2)

, much faster than the near end's

\ell/x_0

❌ Not quite. Write the exact field divided by the point-charge field, set it to

0.99

(or

1.01

) and solve.

Show solution

(i) Dividing the exact field by $Q/(4\pi\varepsilon_0 x_0^2)$ :

\frac{E_{\text{exact}}}{E_{\text{point}}} = \frac{x_0}{x_0+\ell} = 0.99 \;\Rightarrow\; x_0 = 0.99\,x_0 + 0.99\,\ell \;\Rightarrow\; x_0 = 99\,\ell

(ii) With $d = x_0 + \ell/2$ , substitute $x_0 = d - \ell/2$ into the on-axis result: $x_0(x_0+\ell) = (d - \ell/2)(d + \ell/2) = d^2 - \ell^2/4$ , so the exact field is $Q/\big(4\pi\varepsilon_0(d^2-\ell^2/4)\big)$ , slightly stronger than $Q/(4\pi\varepsilon_0 d^2)$ :

\frac{E_{\text{exact}}}{E_{\text{point}}} = \frac{d^2}{d^2 - \ell^2/4} = 1.01 \;\Rightarrow\; 0.01\,d^2 = 1.01\cdot\frac{\ell^2}{4} \;\Rightarrow\; d^2 = 25.25\,\ell^2

d \approx 5.02\,\ell

Placing the point charge at the centre is the far better approximation: it is good to $1\%$ at $d \approx 5\ell$ , while the near-end version needs $x_0 = 99\ell$ .

Problem 4 · Doubling the Rod

Given: a uniform rod with density $\lambda$ and length $\ell$ ; $P$ is on its axis, $x_0 = \ell$ from the near end. The rod is extended at its far end to length $2\ell$ , with the same $\lambda$ (so the charge doubles). Find $E_{\text{new}}/E_{\text{old}}$ at $P$ .

✅ Correct! The added charge sits between

2\ell

and

3\ell

from

P

, so doubling the charge raises the field by only a third.

❌ Not every element is equally close. The new charge lies farther from

P

than the old, and each element contributes

\propto 1/x'^2

, so the field less than doubles.

❌ The charge changed too. Keeping

\lambda

fixed while doubling the length doubles

Q

; using

Q/\big(x_0(x_0+\ell)\big)

with the old

Q

gives

2/3

❌ The added charge still contributes. Fields superpose — nothing shields

P

from the far half — so every positive element adds a

-\hat{i}

contribution.

Show solution

Write the axis field in terms of $\lambda$ (the integral before simplifying):

E = \frac{\lambda}{4\pi\varepsilon_0}\int_{x_0}^{x_0+\ell}\frac{dx'}{x'^2} = \frac{\lambda}{4\pi\varepsilon_0}\left(\frac{1}{x_0} - \frac{1}{x_0+\ell}\right)

Old rod ( $x_0 = \ell$ , length $\ell$ ): $\ \dfrac{1}{\ell} - \dfrac{1}{2\ell} = \dfrac{1}{2\ell}$ .

New rod ( $x_0 = \ell$ , length $2\ell$ ): $\ \dfrac{1}{\ell} - \dfrac{1}{3\ell} = \dfrac{2}{3\ell}$ .

\frac{E_{\text{new}}}{E_{\text{old}}} = \frac{2/(3\ell)}{1/(2\ell)} = \frac{4}{3}

The extra segment, from $x' = 2\ell$ to $3\ell$ , adds $\dfrac{1}{2\ell} - \dfrac{1}{3\ell} = \dfrac{1}{6\ell}$ — one third of the original $\dfrac{1}{2\ell}$ .

Solved: 0 / 4