PHY1002 · Week 10 · Capacitance and Dielectrics Ch 26, An Atomic Description of Dielectrics.

Atomic description of dielectrics and midterm review

Explains why the field inside a dielectric drops by a factor κ using polarization and induced surface charge, then collects the formulas of weeks 1-9 in one table and gives mixed midterm questions.

polarizationpolar moleculenonpolar moleculeinduced dipoleinduced surface chargefree chargebound chargedielectric constantdielectric breakdown

Last week you used, as a rule, that a dielectric increases the capacitance by a factor κ. This week you see why at the molecular level: the material is polarized, induced charge appears on its surfaces, and this charge partly cancels the field of the plates. The topic is short and conceptual; on the exam it usually comes as one conceptual question and one induced-charge calculation. The second half of the page is a midterm review: the formulas of weeks 1-9 in one table, and mixed questions.

Golden rules:

A dielectric is not a conductor. Charges do not flow freely inside the material; they only shift a little within each molecule.
Polarization leaves charge on the surface. Inside, the charges of neighboring dipoles cancel; the surfaces stay charged.
The induced charge has the opposite sign to the charge on the plate and is smaller than it: $\sigma_{ind} = \sigma(1 - 1/\kappa)$ .
The net field is $E = E_0/\kappa$ . The larger κ is, the more easily the material is polarized.

Polar and nonpolar molecules

Molecules respond to an electric field in two different ways.

Polar molecule. Its charge centers are separated even without a field; it has a permanent dipole moment. Water ( $p \approx 6.2\times10^{-30}$ C·m) is the best-known example: the oxygen side is slightly negative, the hydrogen side slightly positive. With no field, thermal motion orients the dipoles randomly and their average is zero. When a field is applied, the torque $\tau = pE\sin\theta$ from week 9 tries to turn them along the field. Thermal motion resists this, so the alignment is partial.

Nonpolar molecule. Symmetric molecules such as $\text{O}_2$ , $\text{N}_2$ , $\text{CO}_2$ have no dipole moment without a field. When a field is applied, the positive nuclei are pushed slightly along the field and the electron cloud slightly the other way. The charge centers separate and an induced dipole forms. Its dipole moment is in the same direction as the field and, for small fields, proportional to it.

TIP · The comb and the paper. This is why a charged comb attracts neutral pieces of paper. The comb’s field induces dipoles in the paper. The end near the comb gets opposite charge, the far end gets charge of the same sign. Because the field falls with distance, the attraction at the near end wins. A nonuniform field can exert a net force on a neutral object.

Induced surface charge

Now put a dielectric slab between the plates of a capacitor charged with $+\sigma$ and $-\sigma$ . The field produced by the plates is $E_0 = \sigma/\varepsilon_0$ , from the positive plate to the negative plate.

Every molecule becomes a dipole in this field (or its existing dipole is aligned). Think of the inside of the slab: the positive end of one molecule touches the negative end of the next. In the interior these cancel. But on the two surfaces there is no neighbor to cancel them:

On the face toward the positive plate the negative ends of the dipoles are left: $-\sigma_{ind}$ .
On the face toward the negative plate the positive ends are left: $+\sigma_{ind}$ .

These two surface charges form their own small capacitor and produce a field in the opposite direction: $E_{ind} = \sigma_{ind}/\varepsilon_0$ . The net field inside the dielectric:

$E = E_0 - E_{ind} = \frac{\sigma - \sigma_{ind}}{\varepsilon_0}$

Since experiment says $E = E_0/\kappa$ :

$\frac{\sigma}{\kappa\varepsilon_0} = \frac{\sigma - \sigma_{ind}}{\varepsilon_0} \quad\Rightarrow\quad \sigma_{ind} = \sigma\left(1 - \frac{1}{\kappa}\right)$

The charge on the plate is called free charge, and the charge on the dielectric surface is called bound charge. Bound charge cannot leave the material; it is tied to its molecule.

κ	$\sigma_{ind}/\sigma$	Comment
1 (vacuum)	0	no polarization
2	0.50	the field is halved
4	0.75	the field falls to a quarter
very large	approaches 1	like a conductor: the field inside goes to zero

The last row is a nice limit: in a conductor the charges flow freely, exactly cancel the surface charge, and the field inside is exactly zero (the electrostatic equilibrium of week 4). A dielectric is a half-finished version of this.

Example 1. On the plates $\sigma = 6.0\ \mu\text{C/m}^2$ , $\kappa = 3.0$ .

E0 = σ/ε0 = 6.0e-6 / 8.85e-12 = 6.78e5 N/C
E  = E0/κ = 2.26e5 N/C
σ_ind = σ(1 − 1/κ) = 6.0 (1 − 1/3) = 4.0 μC/m²
check: (σ − σ_ind)/ε0 = 2.0e-6 / 8.85e-12 = 2.26e5 N/C   (the same)

Example 2. An isolated parallel-plate capacitor: $A = 0.020\ \text{m}^2$ , $d = 1.0\ \text{mm}$ , $Q = 4.0\ \text{nC}$ . A slab with $\kappa = 2.0$ is inserted.

C0 = ε0 A/d = 177 pF        V0 = Q/C0 = 22.6 V
C  = κ C0 = 354 pF           V  = Q/C  = 11.3 V   (Q constant, V halved)
Q_ind = Q(1 − 1/κ) = 2.0 nC  (on each face)

Why did the voltage drop? The free charge is the same but the net field is halved, so $\Delta V = Ed$ is halved too. That is why $C = Q/\Delta V$ doubled. This is the microscopic explanation of the rule $C = \kappa C_0$ from week 9.

PITFALL · Dielectric breakdown. If the field becomes very large, electrons are torn from their molecules and the material becomes a conductor (dielectric breakdown). This limit is the dielectric strength. Don’t confuse polarization with breakdown: in polarization the charge stays in the molecule, in breakdown a current flows.

Midterm review: weeks 1-9

The midterm usually covers topics up to week 9. The table below is everything to look at on the morning of the exam.

Week	Topic	Key formulas	Most common pitfall
1	Coulomb, field of a point charge	$F = k\lvert q_1q_2\rvert/r^2$ , $\vec E = k q\hat r/r^2$	Adding magnitudes instead of adding vectors
2	Continuous distributions, particle in a uniform field	ring axis $E = kQx/(x^2+a^2)^{3/2}$ , $a = qE/m$	Skipping the component that symmetry cancels
3	Flux, Gauss	$\Phi_E = q_{in}/\varepsilon_0$ ; inside a sphere $kQr/R^3$ ; line $2k\lambda/r$ ; plane $\sigma/(2\varepsilon_0)$	Counting the charge outside
4	Conductors	$E = 0$ inside, $E = \sigma/\varepsilon_0$ at the surface	Confusing a conducting and an insulating sphere
5	Potential difference	$\Delta V = -\int\vec E\cdot d\vec s$ , uniform field $\Delta V = -Ed$	Sign: V decreases along the field
6	Potential of a point charge	$V = kq/r$ , $U = kq_1q_2/r$ , $E_x = -\partial V/\partial x$	V is a scalar, add with signs
7	Continuous distributions, conductor	ring axis $V = kQ/\sqrt{x^2+a^2}$ ; conductor equipotential	Thinking V = 0 inside a conductor (it is constant, not zero)
8	Capacitance	$C = Q/\Delta V$ , $\varepsilon_0A/d$ , series/parallel	Using the series and parallel formulas the wrong way round
9	Energy, dielectric, dipole	$U = \tfrac12CV^2$ , $C = \kappa C_0$ , $\tau = pE\sin\theta$	Mixing up battery connected/removed

Problem-solving strategy

First, what is asked? Force and field are vectors, potential and energy are scalars. If it is a vector, split into components; if a scalar, add with signs.
Is there symmetry? Sphere, long line, infinite plane: Gauss’s law. Axis of a ring, disk or rod: the integral formula.
Convert units at the start. Convert cm, mm, nC, μC all to SI, then calculate.
Magnitude check. If you got thousands of newtons between microcoulomb charges, or a speed above the speed of light for an electron, go back.
Watch out for distractors. The wrong choices are made from typical mistakes: $r$ instead of $r^2$ , forgetting ½, using the series and parallel formulas the wrong way round, not converting units. Don’t assume your number is right just because it appears among the choices; check once that you did not make that mistake.

A mixed review example

Example 3. An electron is released from rest at the negative plate of a parallel-plate capacitor whose plates have $\Delta V = 500\ \text{V}$ between them and are $d = 2.0\ \text{cm}$ apart. What are the field, the electron’s acceleration, and its speed when it reaches the positive plate?

Field (uniform):     E = ΔV/d = 500 / 0.020 = 2.5e4 V/m
Force:               F = eE = (1.60e-19)(2.5e4) = 4.0e-15 N
Acceleration:        a = F/me = 4.0e-15 / 9.11e-31 = 4.4e15 m/s²
Speed (by energy):   ½ me v² = eΔV  →  v = √(2eΔV/me) = 1.33e7 m/s
check (kinematics):  v² = 2ad = 2(4.4e15)(0.020) → v = 1.33e7 m/s

Solving the same problem in two ways (week 2 kinematics and week 5 conservation of energy) is the most reliable way to check your answer on the midterm. The speed is about 4 percent of the speed of light; no relativistic correction is needed.

Common mistakes

Wrong idea	The truth
Charge flows inside a dielectric	A dielectric is an insulator; charge shifts in the molecule, it does not flow
A dielectric increases the field	It reduces the field by a factor $\kappa$ : $E = E_0/\kappa$
$\sigma_{ind} = \sigma/\kappa$	$\sigma/\kappa$ is the net charge; $\sigma_{ind} = \sigma(1 - 1/\kappa)$
Nonpolar molecules do not respond to the field	An induced dipole forms
The induced dipole points opposite to the field	The dipole moment is along the field; the surface charge weakens the field
No force ever acts on a neutral object	In a nonuniform field an induced dipole feels a net force
In Gauss’s law the outside charge also counts toward the flux	Only the net charge inside
$E_x = dV/dx$	With a minus sign: $E_x = -dV/dx$

Test yourself

Exam-style questions. Solve them on paper first, then reveal the answer.

Question 1easy

A parallel-plate capacitor has free surface charge density $4.0\ \mu\text{C/m}^2$ on its plates and is completely filled with a dielectric of $\kappa = 4.0$ . What is the magnitude of the induced surface charge density on the dielectric faces?

1.0 μC/m²
3.0 μC/m²
4.0 μC/m²
5.0 μC/m²
16 μC/m²

Show answer

B) 3.0 μC/m²

$\sigma_{ind} = \sigma\left(1 - \dfrac{1}{\kappa}\right) = 4.0\left(1 - \dfrac14\right) = 3.0\ \mu\text{C/m}^2$ .
1.0 μC/m² is the net charge ( $\sigma/\kappa$ ); it is not the induced charge, it is the difference between the charge on the plate and the induced charge.
The induced charge is always smaller than the free charge; 16 is impossible.

Question 2medium

The plates of an isolated parallel-plate capacitor carry free surface charge density $8.0\ \mu\text{C/m}^2$ . The gap is filled with a dielectric with $\kappa = 2.0$ . What is the electric field inside the dielectric?

2.26 × 10⁵ N/C
4.52 × 10⁵ N/C
9.04 × 10⁵ N/C
1.81 × 10⁶ N/C

Show answer

B) 4.52 × 10⁵ N/C

In vacuum it would be: $E_0 = \sigma/\varepsilon_0 = 8.0\times10^{-6}/8.85\times10^{-12} = 9.04\times10^{5}\ \text{N/C}$ .
Inside the dielectric: $E = E_0/\kappa = 4.52\times10^{5}\ \text{N/C}$ . $1.81\times10^{6}$ comes from multiplying by κ. A dielectric reduces the field, it does not increase it.

Question 3easy

A nonpolar molecule is placed in a uniform external electric field. What happens?

Nothing; nonpolar molecules do not respond to fields.
It acquires an induced dipole moment pointing along the field.
It acquires an induced dipole moment pointing opposite to the field.
It gains a net charge.

Show answer

B) It acquires an induced dipole moment pointing along the field.

The field pushes the positive nucleus slightly along the field and the electron cloud slightly the other way. The charge centers separate: an induced dipole.
Since the dipole moment is defined from $-$ to $+$ , it points in the same direction as the field.
The total charge of the molecule stays zero; it does not gain charge.

Question 4medium

Why is the electric field inside a dielectric slab between capacitor plates smaller than it would be in vacuum (same free charge)?

The dielectric conducts and cancels part of the charge on the plates.
Aligned dipoles create induced surface charges whose field opposes the applied field.
The dielectric absorbs electric field energy and converts it to heat.
The dielectric increases the plate separation.

Show answer

B) Aligned dipoles create induced surface charges whose field opposes the applied field.

A dielectric is not a conductor; charges do not flow freely, the molecules are only polarized.
Negative induced charge builds up on the face toward the positive plate and positive induced charge on the face toward the negative plate.
The field of these charges is opposite to the external field, and the net field falls to $E_0/\kappa$ .

Question 5easy

Point charges $+3.0\ \mu\text{C}$ and $-5.0\ \mu\text{C}$ are $15\ \text{cm}$ apart. What is the magnitude of the force between them, and is it attractive or repulsive?

0.90 N, attractive
6.0 N, repulsive
6.0 N, attractive
40 N, attractive
6.0 × 10⁴ N, attractive

Show answer

C) 6.0 N, attractive

$F = k\dfrac{|q_1q_2|}{r^2} = (8.99\times10^{9})\dfrac{(3.0\times10^{-6})(5.0\times10^{-6})}{(0.15)^2} = 5.99\ \text{N} \approx 6.0\ \text{N}$ .
Charges of opposite sign attract.
0.90 N comes from dividing by $r$ instead of $r^2$ .

Question 6easy

What is the magnitude of the electric field $3.0\ \text{cm}$ from a $2.0\ \text{nC}$ point charge?

599 N/C
2.0 × 10³ N/C
2.0 × 10⁴ N/C
6.0 × 10⁴ N/C
2.0 × 10⁷ N/C

Show answer

C) 2.0 × 10⁴ N/C

$E = \dfrac{k|q|}{r^2} = \dfrac{(8.99\times10^{9})(2.0\times10^{-9})}{(0.030)^2} = \dfrac{17.98}{9.0\times10^{-4}} = 2.0\times10^{4}\ \text{N/C}$ .
599 N/C comes from using $1/r$ (that is the form of the potential).

Question 7medium

A closed surface encloses charges $+4.0\ \text{nC}$ and $-1.0\ \text{nC}$ . A $+2.0\ \text{nC}$ charge lies outside the surface. What is the net electric flux through the closed surface?

113 N·m²/C
339 N·m²/C
565 N·m²/C
791 N·m²/C
0

Show answer

B) 339 N·m²/C

Gauss's law: $\Phi_E = q_{in}/\varepsilon_0$ , only the enclosed charge counts. $q_{in} = 4.0 - 1.0 = 3.0\ \text{nC}$ , $\Phi_E = 3.0\times10^{-9}/8.85\times10^{-12} = 339\ \text{N·m}^2/\text{C}$ .
565 comes from ignoring the signs and adding 5.0 nC; 791 includes the outside charge as well. A charge outside does not change the flux; the shape of the surface does not matter either.

Question 8hard

A solid insulating sphere of radius $10\ \text{cm}$ carries a charge of $6.0\ \text{nC}$ distributed uniformly throughout its volume. What is the field magnitude at $r = 5.0\ \text{cm}$ from the center?

0
2.7 × 10³ N/C
5.4 × 10³ N/C
1.1 × 10⁴ N/C
2.2 × 10⁴ N/C

Show answer

B) 2.7 × 10³ N/C

Inside, the charge enclosed by the Gaussian sphere is $q_{in} = Q(r/R)^3 = Q/8$ . $E = \dfrac{kQr}{R^3} = \dfrac{(8.99\times10^{9})(6.0\times10^{-9})(0.050)}{(0.10)^3} = 2.7\times10^{3}\ \text{N/C}$ . $2.2\times10^{4}$ comes from treating all the charge as a point ( $kQ/r^2$ ), and 0 from confusing it with a conductor.

Question 9medium

How much work must an external agent do to bring a $+2.0\ \mu\text{C}$ charge from infinity to a point $30\ \text{cm}$ from a fixed $+5.0\ \mu\text{C}$ charge (slowly, no kinetic energy change)?

−0.30 J
0.090 J
0.30 J
1.0 J
3.0 J

Show answer

C) 0.30 J

$W_{\text{ext}} = \Delta U = \dfrac{kq_1q_2}{r} = \dfrac{(8.99\times10^{9})(2.0\times10^{-6})(5.0\times10^{-6})}{0.30} = 0.30\ \text{J}$ .
Bringing like charges together requires positive work. $-0.30$ J is the work done by the field.
1.0 J comes from dividing by $r^2$ .

Question 10medium

An electron starts from rest and is accelerated through a potential difference of $500\ \text{V}$ . What is its final speed (in m/s)?

Show answer

1.3 × 10⁷ m/s

Conservation of energy: $\tfrac12 m_ev^2 = e\,\Delta V$ . $v = \sqrt{\dfrac{2e\Delta V}{m_e}} = \sqrt{\dfrac{2(1.60\times10^{-19})(500)}{9.11\times10^{-31}}} = \sqrt{1.76\times10^{14}} = 1.33\times10^{7}\ \text{m/s}$ .
If you forget the 2 you get $9.4\times10^{6}$ m/s.

Question 11medium

In a region the electric potential is $V(x) = 3x^2 - 2x$ , with $V$ in volts and $x$ in meters. What is the $x$ -component of the electric field at $x = 1.5\ \text{m}$ (in V/m)?

Show answer

−7.0 V/m

$E_x = -\dfrac{dV}{dx} = -(6x - 2)$ .
At $x = 1.5$ : $E_x = -(9.0 - 2) = -7.0\ \text{V/m}$ . The field points in the $-x$ direction.
Don't forget the minus sign. 3.75 V comes from computing $V$ itself.

Question 12easy

A $10\ \mu\text{F}$ and a $15\ \mu\text{F}$ capacitor are connected in series across a $50\ \text{V}$ source. What is the charge on the $10\ \mu\text{F}$ capacitor?

120 μC
300 μC
500 μC
750 μC
1250 μC

Show answer

B) 300 μC

$C_{eq} = \dfrac{10\cdot15}{25} = 6.0\ \mu\text{F}$ , $Q = 6.0\cdot50 = 300\ \mu\text{C}$ .
In series every capacitor carries the same charge: 300 μC.
500 μC comes from applying the full 50 V to the 10 μF; 1250 μC comes from the parallel sum ( $25\cdot50$ ).

Verification: All numerical answers were computed with node (verify/PHY1002/10.mjs): the induced charge was found with σ_ind = σ(1 − 1/κ) and checked by recomputing the field from the net charge; the field inside the insulating sphere was verified with a Simpson integral of the enclosed charge, and E = −dV/dx with a central-difference numerical derivative; the other review questions were computed directly from the formulas.

Review card

inside a dielectric E = E0/κ (κ ≥ 1, it REDUCES the field) polar molecule: permanent dipole, the field rotates (aligns) it nonpolar molecule: the field creates an induced dipole, p along the field polarization → induced (bound) charge on the surfaces, net 0 in the interior σ_ind = σ (1 − 1/κ) always smaller than σ net field: E = (σ − σ_ind)/ε0 = σ/(κ ε0) a dielectric is NOT a conductor: charge does not flow, it only shifts REVIEW: F = k|q1 q2|/r² E = k|q|/r² V = kq/r U = k q1 q2 / r Φ = q_in/ε0 inside an insulating sphere E = kQr/R³ inside a conductor E = 0 ΔV = −∫E·ds uniform field ΔV = −Ed Ex = −dV/dx ½mv² = |q|ΔV C = Q/ΔV parallel plate ε0A/d series: same Q, parallel: same V U = ½CV² u = ½ε0E² dipole τ = pE sinθ, U = −pE cosθ