University Physics II · Electric Charge and Coulomb's Law · 1.7
Electrostatic Material Response
Put any material in a field and its charge moves — free charge to the surface, bound charge by a fraction of an atomic radius. The near face always ends up oppositely charged, so neutral matter is pulled in, never pushed away.
Build the model
Connect the measurement to the mechanism.
An applied field never leaves matter alone. In a conductor the free charge runs to the surface and keeps going until the field inside the material is zero. In an insulator the bound charge cannot travel, so each atom stretches into a small dipole, p = αE.
Both end with the same geometry: the face nearer the source carries the opposite sign, the far face the same sign. Because the near face also sits in the stronger field, attraction beats repulsion, and neutral matter is pulled toward a charged object whatever that object's sign — with a force falling as r⁻⁵ for a point source, not r⁻². Take the conductor case to its conclusion and you have shielding: a hollow conductor's cavity is field-free because the redistribution runs to completion, while a dielectric only divides the field by κ.
All of it is a small-field expansion, and it fails where the surrounding medium ionizes, long before any atom is strained.
- Simple definition
- Electrostatic material response is what a neutral material does in an applied field: free charge moves to the surfaces, bound charge stretches into induced dipoles, and both leave the near face oppositely charged to the source.
- Example
- Hold a charged rod near an uncharged soda can: charge shifts to the near end, the can rolls toward the rod, and it still rolls toward the rod when the rod's sign is reversed.
Linear because the applied field shifts charge by a minute fraction of an atomic radius.
p in C m; α in C m² V⁻¹; polarizability volume α′ = α/(4πε₀) in m³
Polarizability is set by the geometry, not by how much metal is inside.
R in m; a metallised hollow shell of the same radius has the same α′
Maximum 3ε₀E₀ at the poles, zero at the equator, opposite signs either side of it.
C m⁻²; θ measured from E₀; the sphere's net charge is still zero
The q² erases the source's sign; halve the separation and the pull rises by 2⁵ = 32.
N; the second form is a conducting sphere of radius R at r ≫ R, always attractive
Bound charge can never exceed the free charge, so a dielectric reduces a field but never removes it.
κ dimensionless and ≥ 1: air 1.0006, paper ≈ 3, water 80
Conductor and insulator are two ends of one timescale, not two different mechanisms.
ρ is resistivity in Ω m; copper 1.5 × 10⁻¹⁹ s, a good insulator ~10³ s
A conductor answers with its surface
Put a neutral conductor in a field E₀. Its free charge belongs to no particular atom, so it moves, and it keeps moving as long as any field remains inside the metal — a residual internal field is exactly what would drive current. Equilibrium is the arrangement with zero interior field, and everything the material has done shows up as surface charge. For a sphere that density is σ = 3ε₀E₀ cos θ: negative on the pole facing a positive source, positive on the pole facing away, zero around the equator. Nothing was created — integrate σ over the whole sphere and you get zero. Take a 1.00 mm sphere in E₀ = 1.20 × 10⁶ V m⁻¹. Each hemisphere collects 3ε₀E₀πR² = 1.00 × 10⁻¹⁰ C, about 6.2 × 10⁸ electrons. A solid aluminium sphere that size holds roughly 7.6 × 10²⁰ conduction electrons, so fewer than one in 10¹² has to change hemisphere. The response is large in effect and microscopic in extent.
An insulator answers one atom at a time
Bound charge cannot cross the material, but it can shift inside an atom. The nucleus is pushed one way and the electron cloud the other until the internal restoring force balances the applied field, leaving each atom a small dipole, p = αE. Polarizabilities are atomic in size: as a volume α′ = α/(4πε₀), argon is 1.64 × 10⁻³⁰ m³ and water 1.45 × 10⁻³⁰ m³. The displacement is tiny. Model a hydrogen atom as a nucleus inside a uniform electron ball of radius a₀ = 5.29 × 10⁻¹¹ m; at 3 × 10⁶ V m⁻¹, the field that breaks down dry air, the centres separate by 4πε₀a₀³E/e = 3.1 × 10⁻¹⁶ m — less than a nuclear diameter, six millionths of the atom. That is why p = αE stays linear over any laboratory range. Polar molecules add a second channel: water's permanent dipoles partly align against thermal agitation, which is what raises its static κ to about 80 while its electronic polarizability stays unremarkable.
Why the pull is always toward the source
Both responses share one feature: the near face carries charge opposite in sign to the source, and the near face is also the one in the stronger field. The attraction on it outweighs the repulsion on the far face, so the net force points toward the source. Reverse the source's sign and every induced charge reverses with it, leaving the direction unchanged. Energy says the same thing in one line: an induced dipole has U = −½αE², the half accounting for the work spent stretching it, so the object is driven toward larger E² — toward the source. For a point charge and a small conducting sphere, F = 2kq²R³/r⁵. Take the sphere above, 1.00 mm at 1.50 cm from 30.0 nC, where E₀ = 1.20 × 10⁶ V m⁻¹: F = 2.13 × 10⁻⁵ N, about seventeen times the 1.3 × 10⁻⁶ N weight of a 0.13 mg metallised bead. Move out to 3.00 cm and the pull falls by 2⁵ = 32, to 6.66 × 10⁻⁷ N. Induced attraction is strong and short-ranged.
Shielding is redistribution run to completion
A conductor does not merely reduce the interior field, it removes it, because any leftover field would keep driving charge. Hollow the conductor out and the empty cavity inherits that result: every point of the cavity wall sits at one potential, so no field line can cross the cavity — such a line would start and end on the wall, and integrating E·dl along it would give a potential difference between two points that have none. Whatever you rearrange outside, the cavity stays field-free. That is the Faraday cage; Unit 2 maps the fields properly. Two limits matter already. The shielding is one-way: a charge placed inside the cavity still produces a field outside, because the shell's outer surface must carry the matching charge, unless the shell is grounded and that charge is allowed to leave. And no dielectric reaches completion. The best it does is divide by κ, so even water at κ = 80 leaves one part in 80 of the applied field standing inside.
Conductor and insulator are one timescale apart
How fast does the redistribution finish? Excess charge inside a homogeneous material decays with a time constant τ = ε₀κρ, set by permittivity and resistivity alone. Copper, ρ = 1.7 × 10⁻⁸ Ω m, gives τ = 1.5 × 10⁻¹⁹ s from that continuum formula — shorter than the electron collision time of about 10⁻¹⁴ s, which is what really limits the response, and instantaneous by any electrostatic standard either way. A good insulator with ρ = 10¹⁴ Ω m and κ = 4 gives τ = 3.5 × 10³ s, close to an hour. The two categories are therefore the same physics at different rates, not different physics. That is why charge sprayed onto a plastic rod stays where it lands while a metal sphere equalises before you can measure it, and why a humid day, which adds a conducting surface film and cuts the effective τ, kills a demonstration that worked the day before.
Breakdown, not the atom, sets the ceiling
p = αE is the leading term of an expansion, so it must fail at some field. The natural scale is the field an electron already feels inside an atom, e/(4πε₀a₀²) = 5.1 × 10¹¹ V m⁻¹. Dry air ionizes at about 3 × 10⁶ V m⁻¹, six parts in a million of that; solid dielectrics fail far below the atomic scale too, polyethylene near 2 × 10⁷ V m⁻¹ and mica near 1 × 10⁸ V m⁻¹. So you never reach the nonlinear regime in an electrostatics experiment — the surrounding medium gives out first, and the linear response is safe everywhere it is safe to work. Geometry brings the failure nearer, because surface charge crowds onto regions of small radius of curvature and the field just outside a sharp point runs well above the average. What ends this model in practice is not a strained atom but a spark, and where that spark starts is the next topic's subject.
Change one variable at a time
Make the relationship visible.
Flip the sign slider: the face charges and the dipole arrow both turn round, and the force arrow does not. Then drag r from 1.5 cm to 3.0 cm — the separation only doubles, but the force arrow shrinks by 2⁵ = 32.
FIELD AT THE BEAD E₀1.20 MV m⁻¹
NEAR-FACE CHARGE-100.0 pC
INDUCED PULL F21.31 μN
DIPOLE p / 10⁻¹³ C m1.33
Live interpretationFIELD AT THE BEAD E₀: 1.20 MV m⁻¹. NEAR-FACE CHARGE: −100.0 pC. INDUCED PULL F: 21.31 μN. DIPOLE p / 10⁻¹³ C m: 1.33
Catch the common trap
Explain before calculating.
A neutral conducting bead 1.50 cm from a point charge of +30.0 nC is pulled toward it with 2.13 × 10⁻⁵ N. The point charge is replaced by −30.0 nC and moved to 3.00 cm. What force does the bead feel now?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA neutral conducting sphere of radius 2.00 cm sits in a uniform field of 4.00 × 10⁴ V m⁻¹. State the field inside it, then find the peak induced surface charge density and the total charge gathered on the hemisphere facing the source.
- Free charge keeps moving while any field remains inside the metal, so at equilibrium E = 0 in the interior and everything the sphere has done shows up as surface charge.
- σ = 3ε₀E₀ cos θ is largest at the pole, where cos θ = 1: σₘₐₓ = 3 × 8.85 × 10⁻¹² F m⁻¹ × 4.00 × 10⁴ V m⁻¹ = 1.06 × 10⁻⁶ C m⁻².
- One hemisphere collects Q = 3ε₀E₀πR² = 1.062 × 10⁻⁶ C m⁻² × π × (2.00 × 10⁻² m)² = 1.062 × 10⁻⁶ × 1.2566 × 10⁻³ m² = 1.33 × 10⁻⁹ C.
- That charge is negative on the face toward a positive source, and the far hemisphere carries +1.33 nC — integrate over the whole surface and you still get zero.
AnswerE = 0 inside; σₘₐₓ = 1.06 μC m⁻² at the poles; each hemisphere carries 1.33 nC, negative on the near face.
MediumA metallised bead of radius 2.00 mm and mass 0.500 mg lies 3.00 cm from a point charge of −50.0 nC. Find the induced force on it, compare that with its weight, and say what changes if the source is switched to +50.0 nC.
- R = 2.00 mm is small next to r = 30.0 mm, so the bead responds as one induced dipole and F = 2kq²R³/r⁵.
- Numerator: 2 × 8.99 × 10⁹ N m² C⁻² × (5.00 × 10⁻⁸ C)² × (2.00 × 10⁻³ m)³ = 3.60 × 10⁻¹³ N m⁵.
- Denominator: (3.00 × 10⁻² m)⁵ = 2.43 × 10⁻⁸ m⁵, so F = 3.60 × 10⁻¹³ ÷ 2.43 × 10⁻⁸ = 1.48 × 10⁻⁵ N, directed at the charge.
- Weight = 0.500 × 10⁻⁶ kg × 9.81 m s⁻² = 4.91 × 10⁻⁶ N, so the pull is 1.48 × 10⁻⁵ ÷ 4.91 × 10⁻⁶ = 3.0 times the weight — the bead lifts off and flies to the charge.
- Switching to +50.0 nC changes nothing: q² carries no sign, and every induced charge reverses with the source, so the near face is still the opposite one.
AnswerF = 1.48 × 10⁻⁵ N toward the charge, about 3.0 times the bead's 4.91 × 10⁻⁶ N weight; reversing the source's sign leaves both the size and the direction untouched.
HardA parallel-plate capacitor carries free charge of surface density σf = 25.0 nC m⁻² on its plates, and a sheet of paper (κ = 3.00, ρ = 1.0 × 10¹² Ω m) fills the gap. Find the field in the paper and the bound charge density on its faces, estimate how long a stray blob of free charge inside the paper would take to drain away, and say what a copper sheet would do instead.
- Empty, the gap would hold E₀ = σf/ε₀ = 25.0 × 10⁻⁹ C m⁻² ÷ 8.85 × 10⁻¹² F m⁻¹ = 2.82 × 10³ V m⁻¹.
- The paper divides that by κ: E = E₀/κ = σf/(κε₀) = 25.0 × 10⁻⁹ ÷ (3.00 × 8.85 × 10⁻¹²) = 9.42 × 10² V m⁻¹ — reduced, never removed.
- The reduction is paid for by bound surface charge: σb = σf(1 − 1/κ) = 25.0 × (1 − 0.333) = 16.7 nC m⁻², negative on the face against the positive plate and necessarily smaller than σf.
- Free charge left inside drains with τ = ε₀κρ = 8.85 × 10⁻¹² F m⁻¹ × 3.00 × 1.0 × 10¹² Ω m = 26.6 s — slow enough to watch.
- Copper is the same physics at a different rate: ρ ≈ 1.7 × 10⁻⁸ Ω m gives τ ≈ 1.5 × 10⁻¹⁹ s, and the redistribution runs to completion, so the interior field is 0 rather than E₀/3 — the κ → ∞ end of the same formula.
AnswerE = 9.42 × 10² V m⁻¹ in the paper, σb = 16.7 nC m⁻², τ = 26.6 s; copper would leave E = 0 inside, in about 10⁻¹⁹ s.