Electric field of point charge, dipole, ring, disk, line charge (Gauss's law): Worked examples

Throughout, $k \equiv \frac{1}{4 π ε _{0}} \approx 8.99 \times 1 0^{9} N\cdotm^{2} / C^{2}$ and $ε_{0} \approx 8.85 \times 1 0^{- 12} C^{2} / (N\cdotm^{2})$ .

The scenario matrix

Every problem this topic can throw is one of these cells. Each worked example below is tagged with the cell it fills.

#	Cell (the scenario)	Which formula	Example
A	Point charge, positive, plain distance	$k q / r^{2}$	Ex 1
B	Point charge, negative — direction/sign	$k q / r^{2}$ , direction	Ex 2
C	Two point charges — vector superposition, a component cancels	$\int d E$	Ex 3
D	Dipole, axial vs equatorial compared	$2 k p / r^{3}$ , $k p / r^{3}$	Ex 4
E	Ring, general point + the two limits ( $x = 0$ , $x ≫ R$ ) + max	$k Q x / (R^{2} + x^{2})^{3/2}$	Ex 5
F	Disk general, and its degenerate infinite-sheet limit	$\frac{σ}{2 ε _{0}} [1 - \frac{x}{R ^{2} + x ^{2}}]$	Ex 6
G	Infinite line — real-world word problem	$2 k λ / r$	Ex 7
H	Exam twist: two sheets (capacitor-like), fields add vs cancel	$σ /2 ε_{0}$ superposed	Ex 8

Example 1 — Cell A: a single positive point charge

Worked example Statement

A charge $q = + 5.0 nC$ sits at the origin. Find the field magnitude at $r = 0.20 m$ .

Forecast: small charge, tiny distance — do you expect thousands, or millions, of N/C? Guess the power of ten before computing.

Step 1. Use the point-charge law $E = \frac{k q}{r ^{2}}$ . Why this step? A single point charge is the "atom" — no integration, just plug in. Nothing here breaks the sphere symmetry, so the formula applies directly.

Step 2. Substitute: $q = 5.0 \times 1 0^{- 9} C$ , $r = 0.20 m$ , $r^{2} = 0.040 m^{2}$ . $E = \frac{( 8.99 \times 1 0 ^{9} ) ( 5.0 \times 1 0 ^{- 9} )}{0.040} = \frac{44.95}{0.040} \approx 1124 N/C .$ Why this step? Just arithmetic — but keeping the $1 0^{9}$ and $1 0^{- 9}$ visible shows they cancel, so the answer lands near $1 0^{3}$ , matching a sensible forecast.

Step 3. Direction: radially outward, since $q > 0$ . Why this step? Sign of $q$ sets direction; magnitude alone is an incomplete answer for a vector field.

Verify: Units: $\frac{N\cdotm ^{2} / C ^{2} \cdot C}{m ^{2}} = N/C$ . ✓ Order of magnitude $\sim 1 0^{3} N/C$ matches forecast.

Example 2 — Cell B: a negative charge (sign & direction)

Worked example Statement

Charge $q = - 3.0 nC$ is at the origin. What is the field at a point $P$ located $0.10 m$ to the right ( $+ x$ ) of it? Give magnitude and direction.

Forecast: magnitude only depends on $∣ q ∣$ — but which way does the arrow at $P$ point?

Step 1. Magnitude uses $∣ q ∣$ : $E = \frac{k ∣ q ∣}{r ^{2}}$ . Why this step? The formula's $q$ carries a sign, but "how strong" is set by size, not sign. Splitting size from direction avoids losing the minus sign in arithmetic. $E = \frac{( 8.99 \times 1 0 ^{9} ) ( 3.0 \times 1 0 ^{- 9} )}{( 0.10 ) ^{2}} = \frac{26.97}{0.010} \approx 2697 N/C .$

Step 2. Direction: field of a negative charge points toward it. Why this step? Field lines terminate on negative charge. $P$ is on the $+ x$ side, so the field there points back toward the origin, i.e. in $- x$ .

Verify: If we had used the point-charge test-charge idea — a hypothetical $+ 1 C$ at $P$ would be pulled toward the negative charge — the force (hence field) is $- x$ . ✓ Magnitude $\approx 2.7 \times 1 0^{3} N/C$ .

Example 3 — Cell C: two charges, a component cancels

Figure — Electric field of point charge, dipole, ring, disk, line charge (Gauss's law)

Worked example Statement

Two charges $+ q$ and $+ q$ ( $q = 4.0 nC$ ) sit on the $x$ -axis at $x = \pm 0.30 m$ . Find the field at $P = (0, 0.40) m$ on the $y$ -axis.

Forecast: by symmetry, will the field at $P$ point along $x$ , along $y$ , or diagonally? Decide before computing.

Step 1. Each charge is a distance $r = 0.3 0^{2} + 0.4 0^{2} = 0.50 m$ from $P$ (a 3-4-5 triangle — see figure). Why this step? The distance sets each individual magnitude via $k q / r^{2}$ ; the clean 3-4-5 keeps arithmetic honest.

Step 2. Magnitude of each: $E_{1} = \frac{k q}{r ^{2}} = \frac{( 8.99 \times 1 0 ^{9} ) ( 4.0 \times 1 0 ^{- 9} )}{0.25} = 143.8 N/C .$ Why this step? Both charges are equal and equidistant, so $E_{1} = E_{2}$ — this equality is exactly what makes a component cancel.

Step 3. Resolve. The $x$ -components are equal and opposite → cancel. The $y$ -components add. The $y$ -fraction is $cos θ = \frac{0.40}{0.50} = 0.80$ (see the red arrow's vertical shadow in the figure). Why this step? This is superposition's "silent hero" $\overset{r}{^}$ : choose the axis where symmetry does the work, so only one component survives. $E_{P} = 2 E_{1} cos θ = 2 (143.8) (0.80) = 230.1 N/C, along + y .$

Verify: Direction is straight up ( $+ y$ ) — matches the "mirror symmetry across the $y$ -axis kills the horizontal part" forecast. Units N/C. ✓

Example 4 — Cell D: dipole axial vs equatorial

Worked example Statement

A dipole has $p = 6.0 \times 1 0^{- 30} C\cdotm$ (a water-molecule-scale value). Find the far field at $r = 1.0 \times 1 0^{- 9} m$ on the axis and on the equator, and confirm the axial field is exactly twice the equatorial.

Forecast: which is larger — axial or equatorial? By how much?

Step 1. Axial: $E_{a x} = \frac{2 k p}{r ^{3}}$ . Why this step? On the axis the near charge dominates and the two fields point the same way, so the factor is $2$ . We use $r^{3}$ (not $r^{2}$ ) because the two nearly-equal charges cancel their leading $1/ r^{2}$ terms — the parent note's key dipole fact. $E_{a x} = \frac{2 ( 8.99 \times 1 0 ^{9} ) ( 6.0 \times 1 0 ^{- 30} )}{( 1.0 \times 1 0 ^{- 9} ) ^{3}} = \frac{1.0788 \times 1 0 ^{- 19}}{1.0 \times 1 0 ^{- 27}} \approx 1.079 \times 1 0^{8} N/C .$

Step 2. Equatorial: $E_{e q} = \frac{k p}{r ^{3}}$ (half of axial). Why this step? On the bisector the axial components add but only carry a factor $1$ (not $2$ ), and they point opposite to $p$ (see figure). Same $r^{3}$ law. $E_{e q} = \frac{( 8.99 \times 1 0 ^{9} ) ( 6.0 \times 1 0 ^{- 30} )}{1.0 \times 1 0 ^{- 27}} \approx 5.394 \times 1 0^{7} N/C .$

Step 3. Ratio $E_{a x} / E_{e q} = 2$ exactly. Why this step? Confirms the structural fact — it's not a coincidence of these numbers, it's the $2$ vs $1$ in the formulas.

Verify: $1.079 \times 1 0^{8} /5.394 \times 1 0^{7} = 2.00$ . ✓ Axial wins, forecasted correctly.

Example 5 — Cell E: ring — general point, both limits, and the max

Worked example Statement

A ring of radius $R = 0.10 m$ carries $Q = 8.0 nC$ . Find $E$ (a) at $x = 0.20 m$ on the axis, (b) at the centre $x = 0$ , (c) as $x \to \infty$ (show it becomes a point charge), and (d) locate the axial position of maximum field.

Forecast: will the centre give the biggest field (you're closest to all the charge) — or zero?

Step 1 (a). $E = \frac{k Q x}{( R ^{2} + x ^{2} ) ^{3/2}}$ with $R^{2} + x^{2} = 0.01 + 0.04 = 0.05$ . Why this step? Only the axial component of each $d q$ survives; every $d q$ shares the same $cos θ = x / R^{2} + x^{2}$ , so it pulls out of the integral, leaving this closed form. $(R^{2} + x^{2})^{3/2} = 0.0 5^{1.5} = 0.011180, E = \frac{( 8.99 \times 1 0 ^{9} ) ( 8.0 \times 1 0 ^{- 9} ) ( 0.20 )}{0.011180} \approx 1286 N/C .$

Step 2 (b). At $x = 0$ : the numerator has a factor $x = 0$ , so $E = 0$ . Why this step? By symmetry every radial pull from opposite sides of the ring cancels — closeness loses to symmetry. This is the exact mistake the parent note warns about.

Step 3 (c). For $x ≫ R$ : $(R^{2} + x^{2})^{3/2} \to x^{3}$ , so $E \to k Q x / x^{3} = k Q / x^{2}$ — a point charge. Why this step? From far away the ring's radius is invisible; the field must reduce to the atom-of-everything law. A good limit check.

Step 4 (d). Maximum at $x = R / 2 = 0.10/ 2 \approx 0.0707 m$ (from setting $d E / d x = 0$ ; see the peak in the figure). Why this step? The field grows from $0$ at the centre, peaks, then falls off — so the max sits at an interior point, not at the centre.

Verify (d): Plugging $x = R / 2$ : field there $\approx 2769 N/C$ , larger than the $x = 0.20$ value — consistent with a peak nearer the ring. ✓

Example 6 — Cell F: disk, general and degenerate (infinite sheet)

Worked example Statement

A disk of radius $R = 0.050 m$ has $σ = 2.0 \times 1 0^{- 6} C/m^{2}$ . Find $E$ (a) at $x = 0.050 m$ on the axis, and (b) the limit $x ≪ R$ (infinite-sheet value).

Forecast: at $x = R$ (halfway "out" relative to the disk size) — will the field be near the full sheet value $σ /2 ε_{0}$ , or noticeably less?

Step 1 (a). $E = \frac{σ}{2 ε _{0}} [1 - \frac{x}{R ^{2} + x ^{2}}]$ . Why this step? The disk is nested rings; integrating the ring result gives this bracket. The $[\dots]$ is the "finite-size correction" that a real disk carries and an infinite sheet loses. $\frac{σ}{2 ε _{0}} = \frac{2.0 \times 1 0 ^{- 6}}{2 ( 8.85 \times 1 0 ^{- 12} )} = 1.130 \times 1 0^{5} N/C .$ $\frac{x}{R ^{2} + x ^{2}} = \frac{0.050}{0.0025 + 0.0025} = \frac{0.050}{0.0707} = 0.7071.$ $E = 1.130 \times 1 0^{5} (1 - 0.7071) = 1.130 \times 1 0^{5} (0.2929) \approx 3.310 \times 1 0^{4} N/C .$

Step 2 (b). As $x ≪ R$ : $\frac{x}{R ^{2} + x ^{2}} \to 0$ , bracket $\to 1$ , so $E \to \frac{σ}{2 ε _{0}} = 1.130 \times 1 0^{5} N/C$ . Why this step? Close to a big disk you can't tell it's finite — it looks infinite. WHY constant: backing away reveals more sheet, which cancels the $1/ r^{2}$ falloff exactly.

Verify: At $x = R$ the field ( $3.3 \times 1 0^{4}$ ) is only $\approx 29%$ of the sheet value ( $1.13 \times 1 0^{5}$ ) — so the forecast "noticeably less" is right; edge effects matter once $x$ is comparable to $R$ . ✓

Example 7 — Cell G: infinite line — real-world word problem

Worked example Statement

A long straight power-line wire carries a linear charge density $λ = 3.0 \times 1 0^{- 7} C/m$ . A bird lands $r = 2.0 m$ away. What field magnitude does it sit in, and which direction does the field point?

Forecast: compared to a point charge's $1/ r^{2}$ , a line's $1/ r$ falls off slower. Should the field feel weaker or stronger than "your intuition from point charges" at a few metres?

Step 1. Use Gauss's cylindrical result $E = \frac{λ}{2 π ε _{0} r} = \frac{2 k λ}{r}$ . Why this step? Cylindrical symmetry lets $E$ leave the flux integral (only the curved wall contributes) — so we get a clean formula without integrating, unlike a ring. $E = \frac{2 ( 8.99 \times 1 0 ^{9} ) ( 3.0 \times 1 0 ^{- 7} )}{2.0} = \frac{5394}{2.0} \approx 2697 N/C .$

Step 2. Direction: radially outward, perpendicular to the wire (since $λ > 0$ ). Why this step? The field of a line points away from the line along the shortest (perpendicular) path — the same direction the Gaussian cylinder's curved normal points.

Verify: Halving the distance to $1.0 m$ should double $E$ (because $1/ r$ , not $1/ r^{2}$ ): $2 \times 2697 = 5394 N/C$ — that's exactly $2 k λ /1.0$ . ✓ Slower falloff confirmed.

Example 8 — Cell H: exam twist — two sheets (capacitor-like)

Worked example Statement

Two large parallel sheets carry $+ σ$ and $- σ$ with $σ = 4.0 \times 1 0^{- 6} C/m^{2}$ . Find the field (a) between the sheets and (b) outside them.

Forecast: the "twist" — do the two single-sheet fields ( $σ /2 ε_{0}$ each) add or cancel between the plates? Outside?

Step 1. Each isolated sheet makes $E_{1} = \frac{σ}{2 ε _{0}}$ pointing away from $+ σ$ and toward $- σ$ . Why this step? A single sheet is the degenerate disk limit; its field is uniform, so the whole problem is a superposition of two known uniform fields. $E_{1} = \frac{4.0 \times 1 0 ^{- 6}}{2 ( 8.85 \times 1 0 ^{- 12} )} = 2.260 \times 1 0^{5} N/C .$

Step 2 (a) — between: both fields point the same way (from $+$ toward $-$ ), so they add: $E_{in} = 2 E_{1} = \frac{σ}{ε _{0}} = 4.520 \times 1 0^{5} N/C .$ Why this step? This is the capacitor field $σ / ε_{0}$ — the missing factor of 2 the parent note flags as a common mistake, recovered here by counting both sheets, not both faces of one.

Step 3 (b) — outside: the two fields point in opposite directions and cancel: $E_{out} = E_{1} - E_{1} = 0.$ Why this step? Outside, you're on the far side of both sheets; the $+$ sheet pushes out while the $-$ sheet pulls in — equal magnitudes, opposite signs.

Verify: $E_{in} = σ / ε_{0} = 4.0 \times 1 0^{- 6} /8.85 \times 1 0^{- 12} = 4.520 \times 1 0^{5} N/C$ = $2 \times 2.260 \times 1 0^{5}$ . ✓ And $E_{out} = 0$ , matching an ideal Parallel plate capacitor. ✓

Recall check

Recall Why is the ring's field zero at its centre but not at

$x = R / 2$ ? At the centre ::: every $d q$ has an opposite partner pulling equally the other way, so all field vectors cancel — symmetry beats closeness; the field then rises to a maximum at $x = R / 2$ before decaying as $1/ x^{2}$ .

Recall Between two oppositely-charged sheets the field is

$σ / ε_{0}$ , outside it is what? Zero ::: the two single-sheet fields ( $σ /2 ε_{0}$ each) add inside but cancel outside.

Related: Gauss's law · Coulomb's law · Flux and field lines · Electric dipole in a uniform field · Electric potential.