University of Potsdam Polymer ScienceApplied Condensed Matter Physics Physical & Engineering PropertiesSummer 2012 Problem Set 1

Tutor:Dr. Dmitry Rychkov (Room 2.28.0.003, Tel.: (0331) 977-5452, Email: dmitry.rychkov@uni-potsdam.de)

Topics: Electric Fields, Polarization, Dielectrics

1. A parallel plate capacitor (plate separation: 1.0 cm) is charged in vacuum with a voltage ofV0 = 1600 V, and subsequently disconnected from the power supply.

(a) Calculate the voltage if a metal plate of 3 mm thickness is inserted between the capacitorplates without touching them.

(b) Calculate the voltage if the metal plate is replaced by a glass plate (εr = 6) of the samethickness.

(c) For problem (b), calculate the electric field strength in vacuum and inside the glass plate.

σ

σ2

σ1~E1x

d~E2

ǫr

2. An electret material (dielectric constant εr) contains a pos-itive space charge layer σ at a distance x from the bottomelectrode. The sample is short-circuited and is electricallyneutral, so that compensation charges accumulate on theelectrodes.

(a) Calculate the electric fields ~E1 and ~E2 above and belowthe charge layer. (Hint: use Gauss’ law and the fact that the sample is short-circuited.)

(b) Calculate the compensation charge charge densities σ1 and σ2.

3. The force constant between neighboring atoms in NaCl is36 N/m. Their equilibrium distance is 0.282 nm.

(a) Each ion carries a charge of +e or −e. Calculate the equilibrium dipole moment of an ionpair.

(b) Calculate the change in equilibrium distance and dipole moment when a local electric fieldof 1500 MV/m is applied.

(c) Calculate the static polarizability.

4. Consider a parallel plate capacitor connected to a voltagesource. Let the distance between the plates be d = 1 µm and the applied voltage V = 150 V.The capacitor is filled with amorphous Selenium with a permittivity of εr = 6 and a particleconcentration of n = 3.67× 1028 atoms/m3.

(a) What is the polarization within the Selenium?

(b) What is the surface charge density?

(c) Estimate the polarizability of a selenium atom.

(d) Calculate the local field acting on a selenium atom.

(e) Calculate the induced dipole moment of a selenium atom.

Solutions:Problem 1:The three situations which will be encountered are shown in the drawing. The initial vacuum capacitorwas charged to Q1, has a capacitance of C1 = ε0A/l, and experiences a voltage drop of V1 = Q1/C1.

l t

l1

l2

1: , ,Q V C11 1 2: , ,Q V C22 2 3: , ,Q V C33 3

(a) When the metal plate is inserted, mirror charges will appear on it, as shown in the drawing. Thecharge on the outer plates is constant, since they are isolated, therefore Q1 = Q2.

The voltage does not remain constant, neither does the capacitance. The capacitance is situation2 is calculated by treating the system as a series connection of two capacitors,

1

C2=

1

C(l1)+

1

C(l2)=

l1ε0A

+l2ε0A

=l1 + l2ε0A

=l − tε0A

where A is the (arbitrary) area of the capacitors. The sum of the separation lengths is equal to thedistance between outer electrodes minus the thickness of the inserted metal plate, l1 + l2 = l− t.The new voltage is then found from

Q1 = Q2 ⇔ C1V1 = C2V2 ⇔ V2 = V1C11

C2= V1 ·

ε0A

l· l − tε0A

= V1l − tl

= V17mm

10mm

which evaluates to 1120 V.

(b) Again we note that the charge on the outer plates must be conserved. The capacitor has changed,we now consider it to be a series connection of three capacitors:

1

C3=

1

C(l1)+

1

C(t)+

1

C(l2)=

l1ε0A

+t

εrε0A+

l2ε0A

=l − tε0A

+t

εrε0A

and the voltage becomes

Q1 = Q3 ⇔ V3 = V1C11

C3= V1

l − t+ tεr

l= V1

7mm + 3mm6

10mm= 3

4V1

which is 1200 V.

(c) Again, since the capacitor plates are isolated, the amount of charge is unchanged, and thereforealso the displacement, D = Q/A is unchanged. Further, we know that the displacement onlychanges due to free charges, therefore the insertion of a dielectric plate does not alter the amountof displacement inside the capacitor, D1 = D3, see drawing. For clarification, the dielectric glassplate was moved to lie against the capacitor plate, which does not change the problem.

3: , ,Q V C33 3

a l-t:

b t:

V D Ea a a, ,

V D Eb b b, ,

The total voltage drop from one plate to the other was calculated in the previous question. Thisdrop can be split into two parts, the drop over the vacuum, Va, and the drop inside the dielectric,Vb, such that V3 = Va + Vb. We need to find both Va and Vb to extract the electric fields.

Da = Db ⇔ ε0Ea = εrε0Eb ⇔ Val − t

= εrVbt

⇔ Va = Vb εrl − tt

= V3 − Vb

Isolation of Vb yields

Vb =V3

1 + εrl−tt

= 80V =⇒ Eb =Vbt

= 26.67kV

m

We can now calculate the voltage drop in the vacuum,

Va = V3 − Vb = 1120V =⇒ Ea =Val − t

= 160kV

m

We could also calculate the field from noting that the electric field inside the dielectric is reducedby an amount proportional to the dielectric constant, Eb = Ea/εr, such that

Ea = εrEb = 160kV

m

the same number.

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(d) Bonus question: (d) If the capacitor remains connected to the power supply, how does its chargechange if either the metal plate or the glass plate is inserted between the capcaitor plates?

Answer: A similar approach as previously, except now the voltages remain constant, V1 = V2 = V3.For the insertion of the metal plate:

Q1

C1=Q2

C2=⇒ Q2

Q1=C2

C1=

(ε0Al−t

)(ε0Al

) =l

l − t=

10

7

The amount of charges increases by an amount of 37Q when inserting a metal plate.

A similar approach for the dielectric plate yields

Q3

Q1=C3

C1=

(ε0A

l−t+ tεr

)(ε0Al

) =l

l − t+ tεr

=4

3

The amount of charges increases by an amount of 13Q when inserting a glass dielectric with ε = 6.

Problem 2:We start by putting a Gauss box around the charge layer inside the electret:

1

4 3

2

E1

E2 dA

dA

dA

dA

Gauss’ Law states that the total displacement on the surface of a closed volume equals the total freecharge inside that volume,

∮~D · d ~A = Q.

(a) We treat each surface in the above drawing separately:

1 :

∫1

~D · d ~A = εrε0

∫1

~E · d ~A = −εrε0E1A

2 :

∫2

~D · d ~A = εrε0

∫2

~E · d ~A = εrε0E2A

3 :

∫3

~D · d ~A = 0

4 :

∫4

~D · d ~A = 0

=⇒∮~D · d ~A = −εrε0E1A+ εrε0E2A+ 0 + 0 = σ A ⇔ σ = εrε0(E2 − E1)

The short circuit condition requires that the total potential drop from top to bottom is zero,V1 + V2 = 0. The electric field below the charged layer is connected to the voltage drop in thispart thorugh E1 = V1/x, and above E2 = V2/(d − x), such that the short circuit condition maybe written as

V1 + V2 = 0 = E1 · x+ E2 · (d− x) = E1x+

(σ

εrε0+ E1

)(d− x)

⇔ E1 = − σ

εrε0

(1− x

d

)E2 =

σ

εrε0+ E1 =

σ

εrε0

x

d

Notice that E1 is negative and E2 positive.

(b) The compensation charges are also found with the help of Gauss’ law,

Page 3

0

00

0 0

0

On the lower plate,

σ1A =

∫εrε0 ~E · d ~A = εrε0E1A

⇔ σ1 = εrε0

(− σ

εrε0

(1− x

d

))= σ

(1− x

d

)

and on the upper plate,

σ2 = −εrε0σ

εrε0

x

d= −σ x

d

Notice that both the charge densities found have signs opposite to the charge in the center.

Problem 3:(a) The electric dipole moment is µ = qd = er0 = 4.52 · 10−29 Cm.

(b) Assuming that the dipole is aligned with the electric field lines, there will be a total force actingon the dipole given by F = 2qE = 2eE. This force is balanced by the elasticity of the problem,F = k ·∆x,

F = 2eE = k∆x ⇔ ∆x =2eE

k= 1.335 · 10−11m = 13.35pm

(c) The change in polarisation due to the application of an electric field is proportional to the polar-isability, ∆P = αE. In this case, the polarisability dipole change can be calculated directly fromthe change in distance between the charges in the molecule,

∆P = ∆x · q = ∆x · e =2eE

k· e =

2e2

k· E =⇒ α =

2e2

k= 1.43 · 10−39Cm2

N

Problem 4:(a) The polarisation is found from the electric field and dielectric constant by

P = (εr − 1)ε0E = 5V

d= 6.64 · 10−3 C

m2

(b) The surface charge density is also known as the displacement,

D = εrε0E = 6V

d= 7.97 · 10−3 C

m2

(c) Aided by the Clausius–Mosotti:

εr − 1

εr + 2=nα

3ε0=⇒ α =

3ε0n

εr − 1

εr + 2=

15ε08n

= 4.52 · 10−40 C

Vm

(d) The necessary equation was given by Blythe and Bloor:

EL =εr + 2

3E =

8V

3d= 400

V

m

(e) The dipole moment induced by an applied electric field is found from the polarisability

µ = αE = 6.785 · 10−32Cm

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