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

True or false — justify

A charged body with zero net charge produces zero electric field everywhere.

False. A dipole has net charge zero yet its field is nonzero (dies as

1/ r^{3}

); only the far-field average washes out, not the field itself.

Gauss's law

\oint E \cdot d A = Q_{e n c} / ε_{0}

holds for a ring of charge.

True — it always holds. What fails for a ring is solving it: no surface makes

E

constant and perpendicular, so the integral won't collapse to

E \cdot A

The flux through a closed surface depends on the exact position of charges inside it.

False. Flux depends only on the total enclosed charge; move a charge around inside and the net flux is unchanged, even though the local field pattern shifts. See Flux and field lines.

A charge sitting outside a closed surface contributes zero net flux through it.

True. Its field lines enter one side and exit the other, so inflow and outflow cancel exactly —

Q_{e n c} = 0

for that charge.

The field at the center of a uniformly charged ring is zero.

True. Every element's pull is cancelled by the element diametrically opposite it; symmetry beats "closeness."

The field of an infinite sheet is

σ / ε_{0}

False. A single isolated sheet gives

σ /2 ε_{0}

; the flux exits both faces, so each face carries half. The full

σ / ε_{0}

only appears between two opposite sheets (a Parallel plate capacitor).

Doubling the distance from an infinite line charge halves the field.

True. Line field

\propto 1/ r

, so

r \to 2 r

gives

E \to E /2

— unlike a point charge where it would quarter.

The dipole axial field is exactly twice the equatorial field at the same distance.

True (in the far-field

r ≫ d

E_{a x} = 2 k p / r^{3}

and

E_{e q} = k p / r^{3}

, and they point in opposite senses relative to

p

Spot the error

"For a ring, since I'm closest to all the charge at the center,

E

is maximum there."

The pulls from opposite elements cancel at the center, giving

E = 0

. The maximum is at

x = \pm R / 2

along the axis.

"The dipole is two point charges, each

1/ r^{2}

, so the dipole field is

1/ r^{2}

The leading

1/ r^{2}

terms subtract out because the charges are opposite; expanding the difference leaves a

1/ r^{3}

leading term.

"Gauss's law gives

E

for any shape because it's always true."

True but useless for solving unless symmetry lets

E

leave the integral. Ring, disk, and finite line have no such surface — you must integrate Coulomb's law instead.

"The end caps of the Gaussian cylinder for a line charge contribute flux."

They don't —

E

is radial, so it lies parallel to the flat end caps, making

E \cdot d A = 0

there. Only the curved wall contributes.

"For the disk, as

x \to \infty

the field stays

σ /2 ε_{0}

That's the near-field (

x ≪ R

) limit. Far away the bracket

[1 - x / R^{2} + x^{2}] \to 0

and the disk looks like a point charge,

E \to k Q / x^{2}

"A negative charge makes the field point outward, just weaker."

A negative charge makes the field point inward (toward it). The direction flips with the sign; only the magnitude scales.

E

inside a Gaussian sphere is constant, so I can pull it out — this proves the point-charge law."

The statement is right only because of spherical symmetry around the point charge. If the charge were off-center or non-spherical,

E

would vary on the sphere and could not be pulled out.

Why questions

Why does a point charge's field fall as

1/ r^{2}

Its fixed number of field lines spread over a sphere of area

4 π r^{2}

; line density (= field strength) therefore drops as

1/ r^{2}

. That is Gauss's law in picture form.

Why does the dipole field fall faster (

1/ r^{3}

) than a point charge?

The two opposite charges nearly cancel at large distance; what survives is only their small difference, which shrinks one extra power of

r

faster.

Why does the infinite line fall slower (

1/ r

) than a point charge?

Backing away, you "see" a longer stretch of line contributing; the added charge partly compensates the

1/ r^{2}

single-point falloff, netting

1/ r

Why is the infinite sheet's field independent of distance?

Move away and you take in proportionally more sheet area; the extra charge exactly cancels the

1/ r^{2}

weakening — a perfect trade, so

E

is flat.

Why do only the axial components survive on a ring's axis?

For every element there is a mirror element across the axis; their radial components point oppositely and cancel, leaving only the common axial component (each with the same

cos θ = x / R^{2} + x^{2}

Why choose a sphere for a point charge but a cylinder for a line?

You match the Gaussian surface to the symmetry of the source so

E

is constant and perpendicular on it — spherical symmetry for a point, cylindrical for a line.

Why must we assume

r ≫ d

for the compact dipole formulas

2 k p / r^{3}

and

k p / r^{3}

Only then can we drop the

d^{2} /4

terms and keep the leading behaviour; close up, the exact two-charge expression must be used.

Edge cases

What is

E

on a ring's axis as

x \to \infty

It approaches

k Q / x^{2}

— a point charge, since the ring's size becomes negligible from far away.

What is

E

at the exact center of a charged ring?

Exactly zero, by symmetry — every pull is cancelled by the opposite element.

What is the disk field at

x = 0

(right at the surface, on axis)?

σ /2 ε_{0}

— the bracket becomes

1 - 0 = 1

; this is the finite disk's central value, matching the infinite-sheet result.

What happens to the line-charge field as

r \to 0

It diverges (

E \to \infty

), because an idealized line has zero thickness — infinite charge density up close. Real wires have finite radius, so this is a model artifact.

For the equatorial dipole, which way does

E

point relative to

p

Opposite to

p

. On the axis it points along

+ p

; the reversal is a classic sign trap.

At the exact midpoint between

+ q

and

- q

on a dipole's axis, what is the field direction?

Both charges push the same way there (away from

+

, toward

-

), so the field is strong and points from $+$ toward $-$ , i.e. along

- p

inside the gap — opposite to the far-axial direction.

Between two identical parallel sheets of the same sign, what is

E

Zero — the two

σ /2 ε_{0}

fields point oppositely in the gap and cancel. (For opposite signs they add to

σ / ε_{0}

; contrast with Parallel plate capacitor.)

Recall Quick self-test

Dipole field falloff exponent ::: $1/ r^{3}$ Isolated infinite sheet field ::: $σ /2 ε_{0}$ , constant Field at center of a ring ::: zero, by symmetry Does Gauss's law hold for a ring? ::: yes, but it can't be solved by it — no symmetric surface