Applications — sphere, cylinder, infinite plane: Worked examples

This page is the drill ground for Applications of Gauss's Law. The parent note built the three formulas; here we hit every case they can appear in — inside, outside, at the surface, zero-charge regions, limiting distances, negative charge, real-world word problems, and one exam-style twist.

Before we start, one reminder of the vocabulary we will keep using, so no symbol appears unexplained:

Every formula below comes from Gauss's Law applied with the right symmetry; the flux idea itself is Electric Flux.

The scenario matrix

Each row is one class of case the topic can throw at you. The last column names the worked example that covers it.

#	Case class	What makes it tricky	Covered by
A	Sphere, point outside ( $r > R$ )	acts like a point charge	Ex 1
B	Conducting shell, point inside ( $r < R$ )	$E = 0$ , degenerate	Ex 2
C	Solid insulator, point inside ( $r < R$ )	$E \propto r$ , grows from zero	Ex 3
D	Negative charge (sign case)	arrows reverse, magnitude same	Ex 4
E	Line / cylinder, radial field	$1/ s$ fall-off	Ex 5
F	Infinite sheet + conductor surface	$/2 ε_{0}$ vs $/ ε_{0}$	Ex 6
G	Two-plate capacitor (superposition)	add inside, cancel outside	Ex 7
H	Limiting behaviour ( $r \to R$ , $r \to \infty$ )	continuity & decay checks	Ex 8
I	Real-world word problem	translate words → $λ, σ, Q$	Ex 9
J	Exam twist: concentric shells	choose surface, count $Q_{e n c}$	Ex 10

Example 1 — Sphere, point outside (Cell A)

Worked example Statement

A metal sphere of radius $R = 0.10 m$ carries $Q = + 3 μ C$ . Find $E$ at $r = 0.30 m$ from the centre.

Forecast: guess before computing — is this bigger or smaller than the field at the surface? (It must be smaller: you moved away.)

Recognise the region. $r = 0.30 > R = 0.10$ , so we are outside. Why this step? The formula you use depends entirely on which side of $R$ you stand — the parent note gave two different results.
Use the outside law $E_{o u t} = \frac{k Q}{r ^{2}}$ . Why this step? From outside, all the charge "looks like" a point at the centre, so the field is the point-charge field.
Plug in: $E = \frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 6} )}{( 0.30 ) ^{2}} = \frac{27000}{0.09} = 3.0 \times 1 0^{5} N/C$ , pointing radially outward (charge is positive).

Verify: Units: $\frac{( N\cdotm ^{2} / C ^{2} ) ( C )}{m ^{2}} = N/C$ ✓. Sanity: at the surface $r = 0.10$ the field would be $\frac{27000}{0.01} = 2.7 \times 1 0^{6}$ , nine times bigger — consistent with $1/ r^{2}$ and our forecast (smaller far away). ✓

Example 2 — Conducting shell, point inside (Cell B, degenerate)

Example 3 — Solid insulator, point inside (Cell C)

Worked example Statement

A uniformly charged insulating ball has $Q = + 3 μ C$ spread through its volume, $R = 0.10 m$ . Find $E$ at $r = 0.05 m$ .

Forecast: unlike the conductor, charge is now spread inside — so some charge is enclosed. Field should be nonzero but less than at the surface.

Only enclosed charge counts. The bag at $r = 0.05$ holds the charge in a smaller ball. Since charge is spread evenly, the fraction enclosed is the volume fraction $\frac{r ^{3}}{R ^{3}}$ . Why this step? Volume grows like $r^{3}$ ; uniform density means charge ∝ volume.
Use the inside law $E_{in} = \frac{k Q r}{R ^{3}}$ . Why this step? Enclosed charge $\propto r^{3}$ but the sphere's area $\propto r^{2}$ , so one power of $r$ survives → $E \propto r$ , linear from zero.
Plug in: $E = \frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 6} ) ( 0.05 )}{( 0.10 ) ^{3}} = \frac{1350}{0.001} = 1.35 \times 1 0^{6} N/C$ , radially outward.

Verify: At the surface $r = R = 0.10$ this gives $\frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 6} ) ( 0.1 )}{0.001} = 2.7 \times 1 0^{6}$ , exactly the outside value at $r = R$ — the two formulas agree at the boundary ✓. And $1.35 \times 1 0^{6} < 2.7 \times 1 0^{6}$ as forecast.

Figure — Applications — sphere, cylinder, infinite plane

Look at the graph: the red curve rises straight from the centre (Ex 3, inside insulator), the blue curve falls as $1/ r^{2}$ outside, and they touch at $r = R$ — that touching point is the check we just did.

Example 4 — Negative charge, sign case (Cell D)

Worked example Statement

A point-like sphere carries $Q = - 4 μ C$ . Find the field at $r = 0.20 m$ , and state its direction.

Forecast: magnitude uses the same formula; only the direction differs.

Compute magnitude with $∣ Q ∣$ : $E = \frac{( 9 \times 1 0 ^{9} ) ( 4 \times 1 0 ^{- 6} )}{( 0.20 ) ^{2}} = \frac{36000}{0.04} = 9.0 \times 1 0^{5} N/C$ . Why this step? The formula's $Q$ enters as a magnitude for the size of the field; the sign is handled by direction.
Fix direction from the sign. $Q < 0$ ⇒ field arrows point inward, toward the charge. Why this step? The definition of $E$ : it points the way a positive test charge would be pushed — and a positive charge is pulled toward a negative source.

Verify: Flux through a bag around a negative charge is $\frac{Q _{e n c}}{ε _{0}} < 0$ — negative flux means arrows point into the bag, matching "inward". ✓ Magnitude units N/C ✓.

Example 5 — Line / cylinder, $1/ s$ fall-off (Cell E)

Worked example Statement

A long straight wire has $λ = + 2 \times 1 0^{- 7} C/m$ . Find $E$ at $s = 0.05 m$ .

Forecast: this fades like $1/ s$ , slower than a point's $1/ s^{2}$ .

Choose a coaxial cylinder of radius $s$ , length $L$ . Why this step? Axial symmetry makes $E$ radial and constant on the curved side; the flat end caps contribute nothing.
Line-charge law $E = \frac{λ}{2 π ε _{0} s}$ . Why this step? The length $L$ cancels — proof that the infinite-line result doesn't depend on how tall your bag is.
Plug in: $E = \frac{2 \times 1 0 ^{- 7}}{2 π ( 8.85 \times 1 0 ^{- 12} ) ( 0.05 )}$ . Denominator $= 2 π (8.85 \times 1 0^{- 12}) (0.05) = 2.780 \times 1 0^{- 12}$ , so $E = \frac{2 \times 1 0 ^{- 7}}{2.780 \times 1 0 ^{- 12}} \approx 7.19 \times 1 0^{4} N/C$ , radially outward.

Verify: Double $s$ to $0.10$ and the field should halve (not quarter). Formula gives $\approx 3.60 \times 1 0^{4}$ , exactly half. ✓ Contrast with a point charge, which would drop to a quarter — this confirms the $1/ s$ law.

Example 6 — Sheet vs conductor surface (Cell F)

Example 7 — Two-plate capacitor, superposition (Cell G)

Worked example Statement

Two large parallel plates carry $+ σ$ and $- σ$ with $σ = 1.77 \times 1 0^{- 8} C/m^{2}$ . Find $E$ between the plates and outside them. (This is the parallel-plate result.)

Forecast: between plates the two single-sheet fields point the same way and add; outside they oppose and cancel.

Field of each sheet alone: $\frac{σ}{2 ε _{0}} = 1.0 \times 1 0^{3} N/C$ (from Ex 6). Why this step? Superposition — treat each plate as an isolated sheet, then add the vectors.
Between: both point from $+$ toward $-$ , so $E = \frac{σ}{2 ε _{0}} + \frac{σ}{2 ε _{0}} = \frac{σ}{ε _{0}} = 2.0 \times 1 0^{3} N/C$ .
Outside: the two fields point opposite ways and cancel → $E = 0$ . Why this step? Direction is everything: same direction → add, opposite → subtract.

Verify: $\frac{σ}{ε _{0}} = \frac{1.77 \times 1 0 ^{- 8}}{8.85 \times 1 0 ^{- 12}} = 2.0 \times 1 0^{3}$ ✓, and the "between" value equals the single-conductor value of Ex 6 — makes sense: a capacitor gap behaves like the outside of one conductor. ✓

The arrows show it directly: inside the gap the blue and yellow arrows stack (double strength); outside, they point against each other and vanish.

Example 8 — Limiting behaviour (Cell H)

Worked example Statement

For the solid insulating ball of Ex 3 ( $Q = + 3 μ C$ , $R = 0.10 m$ ), check the two limits: (a) approaching the surface from inside and outside, (b) very far away.

Forecast: the field should be continuous at $r = R$ and vanish as $r \to \infty$ .

From inside, $r \to R$ : $E_{in} (R) = \frac{k QR}{R ^{3}} = \frac{k Q}{R ^{2}} = \frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 6} )}{0.01} = 2.7 \times 1 0^{6} N/C$ . Why this step? Plug $r = R$ into the inside law.
From outside, $r \to R$ : $E_{o u t} (R) = \frac{k Q}{R ^{2}} = 2.7 \times 1 0^{6} N/C$ — identical. Why this step? Same $r = R$ into the outside law; matching values means the graph has no jump.
Far away, $r = 10 m$ : $E = \frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 6} )}{100} = 270 N/C$ , and $\to 0$ as $r \to \infty$ . Why this step? $1/ r^{2}$ decay confirms a finite ball looks like a fading point from far off.

Verify: Inside-limit = outside-limit at $R$ (both $2.7 \times 1 0^{6}$ ) ✓ — continuity confirmed, the same touching point drawn in the Ex 3 figure. And $E (10 m) = 270 ≪ E (R)$ ✓.

Example 9 — Real-world word problem (Cell I)

Worked example Statement

A charged raincloud base is modelled as a huge flat sheet with total charge $- 25 C$ spread over an area of $10 km^{2}$ . Estimate the electric field just below it.

Forecast: we must first turn "total charge over an area" into a $σ$ , then use the sheet formula.

Convert to $σ$ . $A = 10 km^{2} = 10 \times 1 0^{6} m^{2} = 1.0 \times 1 0^{7} m^{2}$ . So $σ = \frac{∣ Q ∣}{A} = \frac{25}{1.0 \times 1 0 ^{7}} = 2.5 \times 1 0^{- 6} C/m^{2}$ . Why this step? The formula needs charge per area, not total charge.
Isolated-sheet field $E = \frac{σ}{2 ε _{0}} = \frac{2.5 \times 1 0 ^{- 6}}{2 ( 8.85 \times 1 0 ^{- 12} )}$ . Why this step? A cloud base seen from just below is effectively an infinite sheet.
Compute: $E = \frac{2.5 \times 1 0 ^{- 6}}{1.77 \times 1 0 ^{- 11}} \approx 1.41 \times 1 0^{5} N/C$ , pointing toward the cloud (charge is negative).

Verify: Units $\frac{C/m ^{2}}{C ^{2} / ( N\cdotm ^{2} )} = N/C$ ✓. Air breaks down (sparks) around $3 \times 1 0^{6} N/C$ ; our $1.4 \times 1 0^{5}$ is below that but large — plausible for a pre-lightning cloud. ✓

Example 10 — Exam twist: two concentric shells (Cell J)

Worked example Statement

An inner conducting sphere ( $R_{1} = 0.05 m$ , charge $+ 2 μ C$ ) sits inside a concentric conducting shell ( $R_{2} = 0.10 m$ , charge $- 2 μ C$ ). Find $E$ at (a) $r = 0.07 m$ (between them) and (b) $r = 0.20 m$ (outside both).

Forecast: the outside field should be zero — the charges cancel. Between, only the inner charge is enclosed.

Between ( $r = 0.07$ ): bag encloses only the inner $+ 2 μ C$ . $E = \frac{k Q _{e n c}}{r ^{2}} = \frac{( 9 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 6} )}{( 0.07 ) ^{2}} = \frac{18000}{0.0049} \approx 3.67 \times 1 0^{6} N/C$ , outward. Why this step? $Q_{e n c}$ counts only charge inside your surface — the outer shell is outside this bag.
Outside ( $r = 0.20$ ): bag now encloses $+ 2 μ C - 2 μ C = 0$ . So $E = 0$ . Why this step? Net enclosed charge is zero → flux zero → field zero.

Verify: Between-value uses only $+ 2 μ C$ ; check $\frac{18000}{0.0049} = 3.673 \times 1 0^{6}$ ✓. Outside, $Q_{e n c} = 0 \Rightarrow E = 0$ — this is why coaxial cables and shields have no external field. ✓

Recall Quick self-test

Field just outside a conductor with surface density $σ$ ::: $E = σ / ε_{0}$ Field inside a hollow conducting shell ::: $E = 0$ Inside a uniformly charged insulating ball, $E$ grows as ::: $E \propto r$ (linearly) Net field outside two equal-and-opposite concentric shells ::: $0$ Between capacitor plates with $\pm σ$ ::: $E = σ / ε_{0}$