Physics 202, Lecture 8Today’s Topics

Capacitance Review Energy storage in capacitors Dielectric materials, electric dipoles Dielectrics and Capacitance

Reminder: Midterm I, Oct 5, 5:30-7:00pm

Exam 1 Reminder

Exam time: Wednesday, October 5, 5:30-7PMCovering: Chapters 21-24, lecture material, lab material(E1 & EC2), and homework.

About 20 Multiple Choice Questions

Bring: Pen/pencil Calculator (no programming functionality!)

1 single-sided formula sheet, self-prepared --no photocopying!

Exam room assignment, special arrangements, otherlogistics, etc will be spelled out in a forthcoming email,which contains also a practice exam.

Capacitors: Summary

• Definition:

• Capacitance depends on geometry:

d

A

- - - - -+ + + +

Parallel Plates

a

b L

r

+Q

-Q

Cylindrical

a

b

+Q-Q

Spherical

C has units of “Farads” or F (1F = 1C/V) ε o has units of F/m

C ≡QΔV

C =εoAd

C = 4πεoabb − a

C =2πεoL

ln ba

⎛⎝⎜

⎞⎠⎟

Capacitors in Series and Parallel

Parallel:

Series:

ΔV

C2

C1

ΔV

Q=Q1+Q2

Cp ΔV=Q

ΔV1 = ΔV2 = ΔV

ΔV1 ΔV2ΔV=ΔV1+ΔV2

Q

ΔV

C1 C2CSΔV=Q

Q1 = Q2 = Q

Cp = C1 + C2

Cs =1C1

+1C2

⎛⎝⎜

⎞⎠⎟

−1

ΔV

Energy of a Capacitor• How much energy is stored in a charged capacitor?

– Calculate the work provided (usually by a battery) to charge acapacitor to +/- Q:

Incremental work dW needed to add charge dq to capacitor at voltage V:- +

• In terms of the voltage V:

• The total work W to charge to Q (W = potential energy of the chargedcapactior) is given by:

Two ways to write W

dW = V (q)idq = q

C⎛⎝⎜

⎞⎠⎟idq

W =1C

qdq =0

Q

∫12Q2

C

W =12CV 2

Capacitor Variables

• In terms of the voltage V:

• The total work to charge capacitor to Q equals the energy Ustored in the capacitor:

You can do one of two things to a capacitor :

• hook it up to a battery specify V and Q follows

• put some charge on it specify Q and V follows

U =

12

CV 2

U =

1C

qdq =0

Q

∫12

Q2

C

Q = CV

V =QC

Example (I)

d

A

- - - - -+ + + +• Suppose the capacitor shown here is charged

to Q. The battery is then disconnected.

• Now suppose the plates are pulled further apart to a finalseparation d1.

• How do the quantities Q, C, E, V, U change?

• Q:• C:• E:• V:• U:

remains the same.. no way for charge to leave.

increases.. add energy to system by separating

decreases.. capacitance depends on geometry

increases.. since C ↓, but Q remains same (or d ↑ but E the same)remains the same... depends only on charge density

C1 =dd1C V1 =

d1dV

U1 =

d1

dU

• Suppose the battery (V) is keptattached to the capacitor.

• Again pull the plates apart from d to d1.

• Now what changes?

• C:• V:• Q:• E:• U:

decreases (capacitance depends only on geometry)must stay the same - the battery forces it to be Vmust decrease, Q=CV charge flows off the plate

d

A

- - - - -+ + + +V

U1 =

dd1

U C1 =

dd1

C E1 =

dd1

E

must decrease ( , ) E =

σE0

E =VD

must decrease ( ) U = 1

2 CV 2

Example (II)

Where is the Energy stored?• Claim: energy is stored in the electric field itself.

• The electric field is given by:

⇒• The energy density u in the field is given by:

Units:

• Consider the example of a constant field generated by a parallelplate capacitor:

++++++++ +++++++

- - - - - - - - - - - - - -- Q

+Q

u =

Uvolume

=UAd

=12ε0 E2

E =σε0

=Qε0A

U =12ε0 E2 Ad

U =

12

Q2

C=

12

Q2

( Aε0 / d)

Jm3

Dielectrics• Empirical observation:

Inserting a non-conducting material (dielectric) between theplates of a capacitor changes the VALUE of the capacitance.

• Definition:The dielectric constant κ of a material is the ratio of thecapacitance when filled with the dielectric to that without it:

κ values are always > 1 (e.g., glass = 5.6; water = 80)Dielectrics INCREASE the capacitance of a capacitorMore energy can be stored on a capacitor at fixed voltage:

κ =CC0

′U =

CV 2

2=κC0V

2

2=κU

ε ≡ κε0permittivity:

Dielectric MaterialsDielectrics are electric insulators:

Charges are not freely movable, but can still have smalldisplacements in an external electric field

Atomic view: composed of permanent (or induced)electric dipoles:

- -

++

Permanent dipole

O--

H+ H+

P

+ - + -

+ - + -

Induced dipole E>0

E=0

+q-qp (dipole moment vector)(see Chapter 21)

Electric Dipole in External E FieldElectric dipole moment p.

Dipole in constant E field:

+

-

d-q

+q

F∑ = 0

τ = p ×

E

U = − piE

Net Force

Net Torque

Potential energy

p = q

d

dipole tends to become aligned with external field!

Dielectrics In External FieldAlignment of permanent dipoles in external field (or alignment

of non-permanent dipoles)

Note: induced field always opposite to the external field E0

Applyingexternal E field

Zeroexternal field Equilibrium

E =E0 −

Eind =

E0κ

E0 =

σε0

Eind =

σ ind

ε0σ ind = σ κ −1

κ

+++++++++++++++

Parallel Plate Example (I)• Deposit a charge Q on parallel plates

filled with vacuum (air)—capacitance C0

• Disconnect from battery• The potential difference is V0 = Q / C0.

Now insert material with dielectric constant κ .

Charge Q remains constantCapacitance increases C = κ C0

Voltage decreases from V0 to:

Q+++++++++++++++

- - - - - - - - - - - - - - -

E0V0

Electric field decreases also:

- - - - - - - - - - - - - - -

Q

V E+- +

-

+- +

-

+-

+-+-

κ

Note: The field onlydecreases when thecharge is held constant!

V =QC

=Q

κC0

=Vκ

E =Vd=V0dκ

=E0κ

++++++++++++++++++

Parallel Plate Example (II)• Deposit a charge Q0 on parallel plates

filled with vacuum (air)—capacitance C0

• The potential difference is V = Q / C0.• Leave battery connected.

Now insert material with dielectric constant κ .

Voltage V remains constantCapacitance increases C = κ C0

Charge increases from Q0 to:

Q0+++++++++++++++

- - - - - - - - - - - - - - -

E0V

Net Electric field stays same (V, d constant)

- - - - - - - - - - - - - - - - - -

Q

V E+- +

-

+- +

-

+-

+-+-

κ

Q = CV =κC0V =κQ0