Electricity-And-Magnetism · Unit 5 · Video 1 · Interactive Practice

A Six-Step Method for the Field of Any Charge Distribution

IKey Formulas

Formula	Step	What it says
$d\vec{\mathbf{E}} = k_e\dfrac{dq}{r^2}\,\hat{\mathbf{r}}$	Steps 1 and 3	Each piece $dq$ is a point charge; $r$ runs from $dq$ to the field point P
$dq = \lambda\,d\ell,\quad dq = \sigma\,dA,\quad dq = \rho\,dV$	Step 2	Charge along a length, over an area, or through a volume
$d\ell = R\,d\phi'$ (ring), $\ dA = 2\pi r'\,dr'$ (disk annulus)	Step 4	Elements written in the coordinates that follow the shape
$E_y = \dfrac{2k_e\lambda}{y}\dfrac{\ell/2}{\sqrt{(\ell/2)^2+y^2}},\quad E_z = \dfrac{k_eQz}{(R^2+z^2)^{3/2}},\quad E_z = 2\pi\sigma k_e\left(\dfrac{z}{\|z\|}-\dfrac{z}{\sqrt{z^2+R^2}}\right)$	Steps 5 and 6	Line (bisector), ring (axis) and disk (axis): only $dE\cos\theta$ survives, then integrate

Key Insight: Only steps 2 (which density) and 4 (which coordinates) change from shape to shape. Step 5, symmetry, does the most work: it turns a vector integral into the single integral of $dE\cos\theta$ . Then check far away, where any finite charge must look like $k_eQ/r^2$ .

IIVisualization 1 — One Pair, Three Shapes

Every piece has a mirror partner, so for the line, the ring and the disk only $2\,dE\cos\theta$ reaches P.

Shape

IIIVisualization 2 — The Far-Away Check

Far from any finite charge the field must approach $k_eQ/z^2$ ; all three results pass that test.

IVQuiz Questions

Problem 1 · Field on the Axis of a Ring

Given: A ring of radius $R = 0.20\ \text{m}$ carries $Q = 5.0\ \text{nC}$ spread uniformly. Point P is on its axis at $z = 0.15\ \text{m}$ . Use $k_e = 9.0\times10^{9}\ \text{N}\,\text{m}^2/\text{C}^2$ . Find $E_z$ at P.

✅ Correct! Every piece sits at

r = \sqrt{R^2+z^2} = 0.25\ \text{m}

and contributes only

dE\cos\theta

with

\cos\theta = z/r = 0.6

❌ Not quite.

k_eQ/r^2 = 720\ \text{N/C}

adds the pieces' magnitudes; their sideways parts cancel, so multiply by

\cos\theta = z/r = 0.6

❌ Not quite.

k_eQ/z^2

treats the ring as a point charge at its centre, which only holds for

z \gg R

; here every piece is

0.25\ \text{m}

away, not

0.15\ \text{m}

❌ Not quite. The surviving factor is

\cos\theta = z/r

, so the numerator carries

z = 0.15\ \text{m}

, not

R

❌ Not quite. Use

E_z = k_eQz/(R^2+z^2)^{3/2}

Show solution

Steps 2–4: $dq = \lambda\,d\ell$ , $d\ell = R\,d\phi'$ , and every piece is at $r = \sqrt{R^2+z^2} = \sqrt{0.04+0.0225} = 0.25\ \text{m}$ . Step 5 keeps $dE\cos\theta$ with $\cos\theta = z/r$ . Step 6 integrates $\phi'$ from $0$ to $2\pi$ , and $2\pi R\lambda = Q$ :

E_z = \frac{k_eQz}{(R^2+z^2)^{3/2}} = \frac{(9.0\times10^{9})(5.0\times10^{-9})(0.15)}{(0.0625)^{3/2}} = \frac{6.75}{0.015625} = 432\ \text{N/C}.

Problem 2 · Field on the Axis of a Disk

Given: A disk of radius $R = 0.10\ \text{m}$ carries uniform surface charge $\sigma = 2.0\times10^{-6}\ \text{C/m}^2$ . Point P is on its axis at $z = 0.10\ \text{m}$ . Use $k_e = 9.0\times10^{9}\ \text{N}\,\text{m}^2/\text{C}^2$ . Find $E_z$ at P.

✅ Correct!

2\pi\sigma k_e = 1.13\times10^{5}\ \text{N/C}

times

1 - 1/\sqrt{2} = 0.293

❌ Not quite.

2\pi\sigma k_e

alone is the infinite-plane field; with

z = R

the second term

z/\sqrt{z^2+R^2} = 0.707

is far from negligible.

❌ Not quite.

k_eQ/z^2

with

Q = \sigma\pi R^2

is the far-away limit; at

z = R

the disk is not yet a point charge.

❌ Not quite. That is

2\pi\sigma k_e

times

z/\sqrt{z^2+R^2}

alone; the bracket is the difference

1 - z/\sqrt{z^2+R^2}

❌ Not quite. Use

E_z = 2\pi\sigma k_e\left(1 - z/\sqrt{z^2+R^2}\right)

for

z > 0

Show solution

Each thin ring has $dA = 2\pi r'\,dr'$ and sits at $r = \sqrt{r'^2+z^2}$ ; integrating $r'$ from $0$ to $R$ gives, for $z > 0$ ,

E_z = 2\pi\sigma k_e\left(1 - \frac{z}{\sqrt{z^2+R^2}}\right) = 2\pi(2.0\times10^{-6})(9.0\times10^{9})\left(1 - \frac{1}{\sqrt{2}}\right).

E_z = (1.131\times10^{5})(0.2929) = 3.3\times10^{4}\ \text{N/C}.

Problem 3 · Field Beside a Line Charge

Given: A line charge of length $\ell = 0.40\ \text{m}$ carries $\lambda = 3.0\ \text{nC/m}$ . Point P is on its perpendicular bisector, $y = 0.15\ \text{m}$ from the midpoint. Use $k_e = 9.0\times10^{9}\ \text{N}\,\text{m}^2/\text{C}^2$ . Find $E_y$ at P.

✅ Correct! The infinite-line value

2k_e\lambda/y = 360\ \text{N/C}

times the end factor

\tfrac{0.20}{0.25} = 0.8

❌ Not quite.

2k_e\lambda/y

is the infinite-line limit; the finite rod adds the factor

\frac{\ell/2}{\sqrt{(\ell/2)^2+y^2}} = 0.8

❌ Not quite.

k_eQ/y^2

with

Q = 1.2\ \text{nC}

treats the rod as a point charge, but P is closer than the rod is long.

❌ Not quite. The factor

2

2k_e\lambda/y

comes from integrating over both halves of the rod; keep it.

❌ Not quite. Use

E_y = \frac{2k_e\lambda}{y}\frac{\ell/2}{\sqrt{(\ell/2)^2+y^2}}

Show solution

With $\ell/2 = 0.20\ \text{m}$ and $y = 0.15\ \text{m}$ , $\sqrt{(\ell/2)^2+y^2} = \sqrt{0.04+0.0225} = 0.25\ \text{m}$ :

E_y = \frac{2(9.0\times10^{9})(3.0\times10^{-9})}{0.15}\cdot\frac{0.20}{0.25} = 360 \times 0.8 = 288\ \text{N/C}.

Problem 4 · Checking the Disk Far Away

Given: the disk result $E_z = 2\pi\sigma k_e\left(1 - \frac{z}{\sqrt{z^2+R^2}}\right)$ for $z > 0$ . Find its form when $z \gg R$ , using $(1+s)^{-1/2} \approx 1 - \tfrac{1}{2}s$ and $Q = \sigma\pi R^2$ .

✅ Correct! The disk passes the far-away check: it looks like a point charge

Q

❌ Not quite.

2\pi\sigma k_e

is the limit close to the disk (

z \ll R

), where it looks like an infinite plane.

❌ Not quite. The expansion gives

1 - z/\sqrt{z^2+R^2} \approx R^2/(2z^2)

; multiplying by

2\pi\sigma k_e

cancels the

\tfrac{1}{2}

❌ Not quite. The bracket tends to

0

, but only like

R^2/(2z^2)

; keep that first-order term.

❌ Not quite. Expand

z/\sqrt{z^2+R^2} = (1 + R^2/z^2)^{-1/2}

to first order.

Show solution

Write $\frac{z}{\sqrt{z^2+R^2}} = \left(1 + \frac{R^2}{z^2}\right)^{-1/2} \approx 1 - \frac{R^2}{2z^2}$ for $z \gg R$ . Then

E_z \approx 2\pi\sigma k_e\cdot\frac{R^2}{2z^2} = \frac{k_e(\sigma\pi R^2)}{z^2} = \frac{k_eQ}{z^2}.

This is step 6's check: far away, a disk of finite size acts as a point charge.

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