Electricity-And-Magnetism · Unit 4 · Video 2 · Interactive Practice

Electric Field on the Perpendicular Bisector of a Charged Rod

IKey Formulas

Formula	Name	What it says
$dE_y = \frac{1}{4\pi\varepsilon_0}\frac{\lambda y\,dx'}{(x'^2 + y^2)^{3/2}}$	One element's $y$ component	$dE_y = dE\cos\theta'$ with $\cos\theta' = y/r'$ ; mirror elements at $\pm x'$ cancel each other's $dE_x$ , so $E_x = 0$
$\int_{-\ell/2}^{\ell/2}\frac{dx'}{(x'^2 + y^2)^{3/2}} = \frac{2\sin\theta}{y^2}$	The integral	$x' = y\tan\theta'$ turns it into $\frac{1}{y^2}\int_{-\theta}^{\theta}\cos\theta'\,d\theta'$
$E_y = \frac{1}{4\pi\varepsilon_0}\frac{2\lambda\sin\theta}{y} = \frac{1}{4\pi\varepsilon_0}\frac{2\lambda}{y}\frac{\ell/2}{\sqrt{y^2 + (\ell/2)^2}}$	Field on the bisector	$\theta$ is the angle at P between the bisector and the line to the rod's end
$y \gg \ell$ : $E_y \approx \frac{Q}{4\pi\varepsilon_0 y^2}$ ; $\ \ell \gg y$ : $E_y \approx \frac{2\lambda}{4\pi\varepsilon_0 y}$	The two limits	A point charge far away; an infinite line of charge close in

Key Insight: Pair before you integrate. Each element alone has an $x$ component, but its mirror image cancels it, so only $dE_y$ needs an integral, and the substitution $x' = y\tan\theta'$ makes each equal step in angle contribute $\cos\theta'\,d\theta'$ .

IIVisualization 1 — Mirror Pairs Cancel $E_x$

Elements at $+x'$ and $-x'$ give equal fields at P whose $x$ components cancel and whose $y$ components add.

IIIVisualization 2 — Where the Field Comes From

In the angle $\theta'$ , each slice of rod contributes $\frac{\lambda}{4\pi\varepsilon_0 y}\cos\theta'\,d\theta'$ , so the distant parts add little.

θ′ out to ± 40.0°

💡 Challenge: find the angle, and the stretch of rod $|x'| \le y\tan\theta'$ , that supplies half of the infinite-line field $\frac{2\lambda}{4\pi\varepsilon_0 y}$ .

IVVisualization 3 — From Line to Point Charge

In units of $E_0 = \frac{Q}{4\pi\varepsilon_0\ell^2}$ , one curve $\frac{E_y}{E_0} = \frac{1}{(y/\ell)\sqrt{(y/\ell)^2 + 1/4}}$ joins both limits.

VQuiz Questions

Problem 1 · Field Beside a Short Rod

Given: A rod of length $\ell = 0.20\ \text{m}$ carries $Q = 4.0\ \text{nC}$ spread uniformly. Point P is on its perpendicular bisector, $y = 0.050\ \text{m}$ from the rod's centre. Use $\frac{1}{4\pi\varepsilon_0} = 9.0\times10^{9}\ \text{N}\,\text{m}^2/\text{C}^2$ . Find $E_y$ at P.

✅ Correct! With

\sin\theta = 0.894

the rod gives

89\%

of the infinite-line value: P is close in, but the ends still matter.

❌ Not quite.

\frac{2\lambda}{4\pi\varepsilon_0 y} = 7.2\times10^{3}\ \text{N/C}

is the infinite-line limit, which assumes

\sin\theta = 1

. Here

\ell/2 = 0.10\ \text{m}

is only twice

y

, so

\sin\theta = 0.894

❌ Not quite.

\frac{Q}{4\pi\varepsilon_0 y^2}

is the far-away limit, valid only for

y \gg \ell

. At

y = 0.050\ \text{m}

P is closer than the rod is long, and that limit overshoots by more than a factor of

2

❌ Check the factor of 2. The integral of

\cos\theta'

runs from

-\theta

+\theta

and gives

2\sin\theta

, so

E_y = \frac{1}{4\pi\varepsilon_0}\frac{2\lambda\sin\theta}{y}

. Using

\cos\theta = 0.447

in place of

\sin\theta

lands on the same wrong value.

Show solution

Line density and end angle:

\lambda = \frac{Q}{\ell} = \frac{4.0\times10^{-9}\ \text{C}}{0.20\ \text{m}} = 2.0\times10^{-8}\ \text{C/m}

\sin\theta = \frac{\ell/2}{\sqrt{y^2 + (\ell/2)^2}} = \frac{0.10}{\sqrt{0.050^2 + 0.10^2}} = \frac{0.10}{0.1118} = 0.894

Field on the bisector:

E_y = \frac{1}{4\pi\varepsilon_0}\frac{2\lambda\sin\theta}{y} = \frac{(9.0\times10^{9})(2)(2.0\times10^{-8})(0.894)}{0.050}\ \text{N/C} \approx 6.4\times10^{3}\ \text{N/C}

It points straight away from the rod (for $Q > 0$ ), and $E_x = 0$ by the pairing argument.

Problem 2 · Adding Magnitudes

Given: A student skips the components and adds magnitudes: $\int_{-\ell/2}^{\ell/2} dE = \frac{1}{4\pi\varepsilon_0}\int_{-\ell/2}^{\ell/2}\frac{\lambda\,dx'}{x'^2 + y^2} = \frac{1}{4\pi\varepsilon_0}\frac{2\lambda\theta}{y}$ (given, with $\theta$ in radians). P sits at $y = \ell/2$ . By what factor does the student's answer exceed the true field?

✅ Correct! The ratio is

\theta/\sin\theta = (\pi/4)/(\sqrt{2}/2) \approx 1.11

. Adding magnitudes counts every element's full

dE

, including the part along

x

that its partner cancels.

❌ Not quite.

1/\cos\theta = \sqrt{2}

is the overcount for the element at the very end only. Elements near the centre are counted almost correctly, so the overall factor is smaller: compare

2\theta

with

2\sin\theta

❌ That is the trap. Cancelling

E_x

means the true field keeps only

dE\cos\theta'

of each element. Summing the full

dE

keeps the

x

parts too, so the student's total is too large.

❌ Check the direction of the ratio.

0.90 = \sin\theta/\theta

is true over student. Since

|dE_y| \le dE

for every element, the student's sum can only be the larger one.

Show solution

At $y = \ell/2$ the end triangle has equal legs, so $\theta = 45\degree = \pi/4$ and $\sin\theta = \sqrt{2}/2$ .

\frac{\text{student}}{\text{true}} = \frac{\frac{1}{4\pi\varepsilon_0}\frac{2\lambda\theta}{y}}{\frac{1}{4\pi\varepsilon_0}\frac{2\lambda\sin\theta}{y}} = \frac{\theta}{\sin\theta} = \frac{\pi/4}{\sqrt{2}/2} = \frac{\pi}{2\sqrt{2}} \approx 1.11

The student's answer is about $11\%$ too large. The error grows as P moves closer (larger $\theta$ ): as $\theta \to 90\degree$ the ratio tends to $\pi/2 \approx 1.57$ .

Problem 3 · When Is the Rod a Point Charge?

Given: The point-charge formula $\frac{Q}{4\pi\varepsilon_0 y^2}$ always overestimates the field on the bisector. Find the distance $y$ at which it overestimates the exact field by exactly $10\%$ .

✅ Correct! Already at about one rod length away, the point-charge formula is within

10\%

❌ Check the square root. The ratio is

\sqrt{1 + (\ell/2y)^2}

. Setting

1 + (\ell/2y)^2 = 1.10

drops the root and asks for a

5\%

error instead.

❌ Check the end triangle. Its horizontal leg is

\ell/2

, not

\ell

: the ratio is

\sqrt{1 + (\ell/2y)^2}

, not

\sqrt{1 + (\ell/y)^2}

❌ Not quite. At

y = \ell/2

the two limit curves cross, but both overestimate there by a factor

\sqrt{2}

, an error of

41\%

Show solution

Divide the point-charge value by the exact field, using $Q = \lambda\ell$ :

\frac{Q/(4\pi\varepsilon_0 y^2)}{\frac{1}{4\pi\varepsilon_0}\frac{2\lambda}{y}\frac{\ell/2}{\sqrt{y^2 + (\ell/2)^2}}} = \frac{\lambda\ell}{y^2}\cdot\frac{y\sqrt{y^2 + (\ell/2)^2}}{\lambda\ell} = \sqrt{1 + \left(\frac{\ell}{2y}\right)^2}

Set it equal to $1.10$ :

\left(\frac{\ell}{2y}\right)^2 = 1.21 - 1 = 0.21 \quad\Rightarrow\quad \frac{\ell}{2y} = 0.458 \quad\Rightarrow\quad y = \frac{\ell}{0.917} \approx 1.09\,\ell

Problem 4 · Half a Rod

Given: Remove the left half of the rod, keeping only the charge from $x' = 0$ to $x' = \ell/2$ (same $\lambda > 0$ ). P stays at height $y$ above the origin, and $\theta$ is still the angle at P to the end $x' = \ell/2$ . Find the field at P.

✅ Correct! The

y

integrand is even in

x'

, so half the rod gives half of

E_y

. Nothing cancels the

x

components, and every one of them points away from the charge, toward

-x

❌ Not quite.

E_x = 0

came from pairing each element with its mirror image at

-x'

. With the left half gone there are no partners, so the

x

components all add.

❌ Check $E_y$ . Only half the elements remain. Since

dE_y

is the same at

+x'

and

-x'

, the remaining half supplies exactly half of

\frac{1}{4\pi\varepsilon_0}\frac{2\lambda\sin\theta}{y}

❌ Check the direction. For

\lambda > 0

each element's field at P points from the element to P. The elements are all at

x' > 0

and P is at

x = 0

, so every

dE_x

is negative.

Show solution

$y$ component. The integrand $\frac{y}{(x'^2 + y^2)^{3/2}}$ is even in $x'$ , so

E_y = \frac{\lambda y}{4\pi\varepsilon_0}\int_{0}^{\ell/2}\frac{dx'}{(x'^2 + y^2)^{3/2}} = \frac{1}{2}\cdot\frac{1}{4\pi\varepsilon_0}\frac{2\lambda\sin\theta}{y} = \frac{1}{4\pi\varepsilon_0}\frac{\lambda\sin\theta}{y}

$x$ component. The element at $x'$ contributes $dE_x = -dE\sin\theta' = -\frac{1}{4\pi\varepsilon_0}\frac{\lambda x'\,dx'}{(x'^2 + y^2)^{3/2}}$ , negative for every $x' > 0$ :

E_x = -\frac{\lambda}{4\pi\varepsilon_0}\left[\frac{1}{y} - \frac{1}{\sqrt{y^2 + (\ell/2)^2}}\right] = -\frac{1}{4\pi\varepsilon_0}\frac{\lambda(1 - \cos\theta)}{y} < 0

Putting the left half back adds an equal $+x$ component, which is how the full rod reaches $E_x = 0$ .

Solved: 0 / 4

Electric Field on the Perpendicular Bisector of a Charged Rod

IKey Formulas

IIVisualization 1 — Mirror Pairs Cancel ExE_xEx​

IIIVisualization 2 — Where the Field Comes From

IVVisualization 3 — From Line to Point Charge

VQuiz Questions

IIVisualization 1 — Mirror Pairs Cancel $E_x$