Electricity-And-Magnetism · Unit 5 · Video 5 · Interactive Practice

Electric Field Off the Axis of a Finite Charged Rod

IKey Formulas

Formula	Name	What it says
$\hat{\mathbf{r}} = -\sin\theta'\,\hat{\mathbf{i}} + \cos\theta'\,\hat{\mathbf{j}},\quad dE_x = -\dfrac{\lambda x'\,dx'}{4\pi\varepsilon_0(x'^2+y^2)^{3/2}},\quad dE_y = \dfrac{\lambda y\,dx'}{4\pi\varepsilon_0(x'^2+y^2)^{3/2}}$	One element	Origin at the foot of the perpendicular from P; no mirror pairs, so both components survive
$E_x = \dfrac{\lambda}{4\pi\varepsilon_0}\left(\dfrac{1}{\sqrt{x_2^2+y^2}} - \dfrac{1}{\sqrt{x_1^2+y^2}}\right) = \dfrac{\lambda}{4\pi\varepsilon_0 y}(\cos\theta_2 - \cos\theta_1)$	Along the rod	From $u = x'^2 + y^2$
$E_y = \dfrac{\lambda}{4\pi\varepsilon_0 y}(\sin\theta_2 - \sin\theta_1)$	Away from the rod	From $x' = y\tan\theta'$
$\theta_1 \to -\tfrac{\pi}{2},\ \theta_2 \to \tfrac{\pi}{2}:\quad E_x = 0,\ E_y = \dfrac{2\lambda}{4\pi\varepsilon_0 y}$	Infinite line	The end angles, measured from the perpendicular, carry all the geometry

Key Insight: An angle is positive when its line leans toward $+x$ . The end farther from the foot of the perpendicular has the smaller cosine, so $E_x$ leans away from the side where the rod extends farther; equal ends make the cosines match and $E_x$ vanish.

IIVisualization 1 — Two End Angles Set the Field

For any point beside the rod, the angles $\theta_1$ and $\theta_2$ to its ends give both components.

IIIVisualization 2 — Growing into an Infinite Line

As a lopsided rod from $-a$ to $2a$ grows, $E_x \to 0$ and $E_y \to \frac{2\lambda}{4\pi\varepsilon_0 y}$ .

a / y 1.00

IVQuiz Questions

Problem 1 · A Rod to One Side: $E_y$

Given: A rod with $\lambda = 5.0\ \text{nC/m}$ runs along the $x$ axis from $x_1 = 0$ to $x_2 = 0.30\ \text{m}$ . Point P is at $(0,\ 0.40\ \text{m})$ . 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!

\theta_1 = 0

\sin\theta_2 = 0.30/0.50 = 0.6

, and

\frac{\lambda}{4\pi\varepsilon_0 y} = 112.5\ \text{N/C}

❌ Not quite.

\cos\theta_2 = 0.8

goes with

E_x

;

E_y

takes the sines.

❌ Not quite.

\frac{2\lambda}{4\pi\varepsilon_0 y}

is the infinite-line limit, with end angles

\pm 90\degree

; here they are

0

and

36.9\degree

❌ Not quite. That is the size of

E_x

, from

\cos\theta_2 - \cos\theta_1

❌ Not quite. Use

E_y = \frac{\lambda}{4\pi\varepsilon_0 y}(\sin\theta_2 - \sin\theta_1)

Show solution

The foot of the perpendicular is the origin, so $\theta_1 = 0$ and $\tan\theta_2 = 0.30/0.40$ : $\sin\theta_2 = 0.6$ , $\cos\theta_2 = 0.8$ .

\frac{\lambda}{4\pi\varepsilon_0 y} = \frac{(9.0\times10^{9})(5.0\times10^{-9})}{0.40} = 112.5\ \text{N/C},\qquad E_y = 112.5\,(0.6 - 0) = 67.5\ \text{N/C}.

Problem 2 · The Same Rod: $E_x$ and Its Sign

Given: the rod and point of Problem 1. Find $E_x$ at P.

✅ Correct! All the charge is to the right of P, so the field leans left:

112.5(0.8 - 1) = -22.5\ \text{N/C}

❌ Check the sign. The order is

\cos\theta_2 - \cos\theta_1 = 0.8 - 1

; each element pushes P away from itself, toward

-x

❌ Not quite. No element has a mirror partner here, so the

x

components do not cancel.

❌ Not quite. Subtract the two cosines:

\cos\theta_2 - \cos\theta_1

, not

\cos\theta_2

alone.

❌ Not quite. Use

E_x = \frac{\lambda}{4\pi\varepsilon_0 y}(\cos\theta_2 - \cos\theta_1)

Show solution

E_x = \frac{\lambda}{4\pi\varepsilon_0 y}(\cos\theta_2 - \cos\theta_1) = 112.5\,(0.8 - 1) = -22.5\ \text{N/C}.

Check with the square-root form: $\frac{\lambda}{4\pi\varepsilon_0}\left(\frac{1}{0.50} - \frac{1}{0.40}\right) = 45(2 - 2.5) = -22.5\ \text{N/C}$ .

Problem 3 · A Rod Straddling the Foot

Given: $\lambda = 5.0\ \text{nC/m}$ on a rod from $x_1 = -0.30\ \text{m}$ to $x_2 = 0.90\ \text{m}$ ; P at $(0,\ 0.40\ \text{m})$ . Use $\frac{1}{4\pi\varepsilon_0} = 9.0\times10^{9}\ \text{N}\,\text{m}^2/\text{C}^2$ . Find $|\vec{\mathbf{E}}|$ at P.

✅ Correct!

E_x = -44.3\ \text{N/C}

and

E_y = 170.3\ \text{N/C}

combine to

176\ \text{N/C}

❌ Not quite. That is

E_y

alone; the rod reaches farther right, so

E_x

is not zero.

❌ Not quite. Combine perpendicular components as

\sqrt{E_x^2 + E_y^2}

, not

|E_x| + E_y

❌ Not quite. The infinite-line value needs end angles of

\pm 90\degree

; here they are

-36.9\degree

and

66.0\degree

❌ Not quite.

\theta_1

is negative here:

\sin\theta_1 = -0.6

Show solution

End 1: $\sqrt{0.09 + 0.16} = 0.50$ , so $\sin\theta_1 = -0.6$ , $\cos\theta_1 = 0.8$ . End 2: $\sqrt{0.81 + 0.16} = 0.985$ , so $\sin\theta_2 = 0.914$ , $\cos\theta_2 = 0.406$ .

E_x = 112.5\,(0.406 - 0.8) = -44.3\ \text{N/C},\qquad E_y = 112.5\,(0.914 + 0.6) = 170.3\ \text{N/C}.

|\vec{\mathbf{E}}| = \sqrt{44.3^2 + 170.3^2} = 176\ \text{N/C}.

Problem 4 · On the Axis as a Limit

Given: A rod with $\lambda = 5.0\ \text{nC/m}$ runs from $x = 0.10\ \text{m}$ to $x = 0.30\ \text{m}$ , and P is at the origin, on the rod's line ( $y \to 0$ ). Use $\frac{1}{4\pi\varepsilon_0} = 9.0\times10^{9}\ \text{N}\,\text{m}^2/\text{C}^2$ . Find $E_x$ at P.

✅ Correct! The square-root form survives

y \to 0

45\,(1/0.30 - 1/0.10) = -300\ \text{N/C}

❌ Check the sign. The rod is on the

+x

side, so a positive rod pushes P toward

-x

❌ Not quite. That puts all of

Q = 1.0\ \text{nC}

at the near end,

0.10\ \text{m}

away; the charge is spread out to

0.30\ \text{m}

❌ Not quite. Concentrating

Q

at the rod's midpoint is not exact for a rod this close; the field is

Q/(4\pi\varepsilon_0\, x_0(x_0 + L))

❌ Not quite. Use

E_x = \frac{\lambda}{4\pi\varepsilon_0}\left(\frac{1}{x_0 + L} - \frac{1}{x_0}\right)

Show solution

In the cosine form $1/y$ blows up while both cosines go to $0$ , so use the square-root form with $y = 0$ , $x_1 = x_0 = 0.10\ \text{m}$ , $x_2 = x_0 + L = 0.30\ \text{m}$ :

E_x = \frac{\lambda}{4\pi\varepsilon_0}\left(\frac{1}{x_0 + L} - \frac{1}{x_0}\right) = 45\left(3.33 - 10\right) = -300\ \text{N/C}.

Equivalently $-\dfrac{Q}{4\pi\varepsilon_0\,x_0(x_0 + L)} = -\dfrac{9.0}{0.030} = -300\ \text{N/C}$ with $Q = \lambda L = 1.0\ \text{nC}$ .

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