1/31/07184 Lecture 141 PHY 184 Spring 2007 Lecture 14.

1/31/07 184 Lecture 14 1

PHY 184PHY 184PHY 184PHY 184

Spring 2007Lecture 14

1/31/07 184 Lecture 14 2

AnnouncementsAnnouncementsAnnouncementsAnnouncements

Midterm 1 will take place in class a week from tomorrow, Thursday, February 8.

One 8.5 x 11 inch equation sheet (front and back) is allowed

The exam will cover Chapters 16 - 19 Homework Sets 1 - 4 Practice exam (the actual Midterm 1 from last fall) is

available: http://www.pa.msu.edu/courses/phy184

1/31/07 184 Lecture 14 3

ReviewReviewReviewReview The electric potential energy stored in a capacitor

is given by

The field energy density stored in a parallel plate capacitor is given by

The field energy density in general is

U 1

2CV 2U

1

2CV 2

u 1

20E

2u 1

20E

2

u 1

20

V

d

2

u 1

20

V

d

2

1/31/07 184 Lecture 14 4

Review (2)Review (2)Review (2)Review (2)

Placing a dielectric between the plates of a capacitor increases the capacitance by (dielectric constant)

The dielectric has the effect of lowering the electric field between the plates (for given charge q)

We also define the electric permitivity of the dielectric material as

C CairC Cair

E Eair

E Eair

0 0

1/31/07 184 Lecture 14 5

Dielectric StrengthDielectric StrengthDielectric StrengthDielectric Strength

The “dielectric strength” of a material measures the ability of that material to withstand voltage differences.

If the voltage across a dielectric exceeds the breakdown potential, the dielectric will break down and begin to conduct charge between the plates.

Real-life dielectrics enable a capacitor provide a given capacitance and hold the required voltage without breaking down.

Capacitors are usually specified in terms of their capacitance and rated voltage.

1/31/07 184 Lecture 14 6

Dielectric ConstantDielectric ConstantDielectric ConstantDielectric Constant

The dielectric constant of vacuum is defined to be 1. The dielectric constant of air is close to 1 and we will use

the dielectric constant of air as 1 in our problems. The dielectric constants of common materials are

Material Dielectric Constant Dielectric Strength (kV/mm)

Air (1 atm) 1.00059 2.5 Polystyrene 2.6 20 Mylar 3.1 280 Paper 3.0 8 Water 80.4 3.1

0

1/31/07 184 Lecture 14 7

Capacitor with DielectricCapacitor with DielectricCapacitor with DielectricCapacitor with Dielectric Question 1: Consider a parallel plate capacitor

with capacitance C = 2.00 F connected to a battery with voltage V = 12.0 V as shown. What is the charge stored in the capacitor?

q CV 2.00 10 6 F 12.0 V 2.40 10 5 C

Question 2: Now insert a dielectric with dielectric constant = 2.5

between the plates of the capacitor. What is the charge on the capacitor?

C Cair The capacitance of the capacitor is increased

C100.6V0.12F100.250.2 56 CVq

The additional charge is provided by the battery.

1/31/07 184 Lecture 14 8

Capacitor with Dielectric (2)Capacitor with Dielectric (2)Capacitor with Dielectric (2)Capacitor with Dielectric (2) Question 3: We isolate the charged capacitor with a dielectric by

disconnecting it from the battery. We remove the dielectric, keeping the capacitor isolated.

What happens to the charge and voltage on the capacitor?

The charge on the isolated capacitor cannot change because there is nowhere for the charge to flow. Q remains constant.

The voltage on the capacitor will be

The voltage went up because removing the dielectric increased the electric field and the resulting potential difference between the plates.

V q

C

6.00 10 5 C

2.00 10 6 F30.0 V V increases

1/31/07 184 Lecture 14 9

Capacitor with Dielectric (3)Capacitor with Dielectric (3)Capacitor with Dielectric (3)Capacitor with Dielectric (3) Question 4: Does removing the dielectric from the isolated

capacitor change the energy stored in the capacitor?

The energy stored in the capacitor before the dielectric was removed was

After the dielectric is removed, the energy is

The energy increases --- we must add energy to pull out the dielectric. [Or, the polarized dielectric is sucked into the E.]

U 1

2CV 2

1

2CairV

2 1

22.50 2.00 10 6 F 12 V 2 3.60 10 4 J

U 1

2CairV

2 1

22.00 10 6 F 30 V 2 9.00 10 4 J

1/31/07 184 Lecture 14 10

ExampleExampleExampleExample

A coaxial cable used in a transmission line has an inner radius of a=0.10 mm and an outer radius of b=0.60 mm. Calculate the capacitance per meter for this type of cable.

Idea: The capacitance of a cylindrical capacitor is given by

The capacitance per unit length is thus:

1/31/07 184 Lecture 14 11

Clicker QuestionClicker QuestionClicker QuestionClicker Question

A coaxial cable used in a transmission line has an inner radius of a=0.10 mm and an outer radius of b=0.60 mm. We calculate the capacitance per meter to be 81 pF/m. Now assume that the space between the conductors is filled with material of =3. How does the capacitance change?

A) 243 pF/m B) 27 pF/m C) doesn’t change D) 8.1 pF/m

pF/m243pF/m813 airCC

1/31/07 184 Lecture 14 12


An air-filled parallel plate capacitor has a capacitance of 1.3 pF. The separation of the plates is doubled, and wax is inserted between them. The new capacitance is 2.6pF. Find the dielectric constant of the wax.

Key Ideas: The original capacitance is given by

Then the new capacitance is

Thus

rearrange the equation:

1/31/07 184 Lecture 14 13


Given a 7.4 pF air-filled capacitor. You are asked to convert it to a capacitor that can store up to 7.4 J with a maximum voltage of 652 V. What dielectric constant should the material have that you insert to achieve these requirements?

Key Idea: The capacitance with the dielectric in place is given by C=Cair

and the energy stored is given by

So,

1/31/07 184 Lecture 14 14

Clicker Question - part 1Clicker Question - part 1Clicker Question - part 1Clicker Question - part 1

A parallel-plate air-filled capacitor has a capacitance of 50 pF.

(a) If each of the plates has an area of A=0.35 m2, what is the separation?

A) 12.5 10-1 m B) 6.2 10-2 m C) 1.3 m

0=8.85 10-12 C2/Nm2

1/31/07 184 Lecture 14 15


A parallel-plate air-filled capacitor has a capacitance of 50 pF.

(a) If each of the plates has an area of A=0.35 m2, what is the separation?

B) 6.2 10-2 m

Use to solve for d:

1/31/07 184 Lecture 14 16


An air-filled parallel plate capacitor has a capacitance of 50pF.

(b) If the region between the plates is now filled with material having a dielectric constant of =2, what is the capacitance?

A) the same B) 25 pF C) 100 pF

1/31/07 184 Lecture 14 17


A air-filled parallel plate capacitor has a capacitance of 50 pF.

(b) If the region between the plates is now filled with material having a dielectric constant of =2, what is the capacitance?

C) 100 pF

1/31/07 184 Lecture 14 18

Microscopic Perspective on DielectricsMicroscopic Perspective on DielectricsMicroscopic Perspective on DielectricsMicroscopic Perspective on Dielectrics

Let’s consider what happens at the atomic and molecular level when a dielectric is placed in an electric field.

There are two types of dielectric materials• Polar dielectric• Non-polar dielectric

Polar dielectric material is composed of molecules that have a permanent electric dipole moment due to their molecular structure• e.g., water molecules

Normally the directions of the electric dipoles are randomly distributed as shown below

1/31/07 184 Lecture 14 19

Microscopic Perspective on Dielectrics (2)Microscopic Perspective on Dielectrics (2)Microscopic Perspective on Dielectrics (2)Microscopic Perspective on Dielectrics (2)

When an electric field is applied to these polar molecules, they tend to align with the electric field

Non-polar dielectric material is composed of atoms or molecules that have no electric dipole moment

1/31/07 184 Lecture 14 20


These atoms or molecules can be induced to have a dipole moment under the influence of an external electric field.

This induction is caused by the opposite direction of the electric force on the negative and positive charges of the atom or molecule, which displaces the center of the relative charge distributions and produces an induced electric dipole moment

1/31/07 184 Lecture 14 21


In both the case of the polar and non-polar dielectric materials, the resulting aligned electric dipole moments tend to partially cancel the original electric field

The electric field inside the capacitor then is the original field minus the induced field

dr EEE

1/31/07 184 Lecture 14 22

Example: Camera FlashExample: Camera FlashExample: Camera FlashExample: Camera Flash

1/31/07 184 Lecture 14 23


1/31/07 184 Lecture 14 24


1/31/07 184 Lecture 14 25

Partially Filled CapacitorPartially Filled CapacitorPartially Filled CapacitorPartially Filled Capacitor A cylindrical capacitor is lying on a

table as shown. The outer cylinder has radius b=4.5 cm, and the inner cylinder has a radius a=4.0 cm. The total length of the capacitor is L=9.0 cm. A dielectric material of constant κ=3.5 is inserted a length L-d=3 cm between the two conducting cylinders. The partially filled capacitor is then connected to a voltage 100.0 V. In what follows, we'll refer to the region where there is no dielectric as region I and the region where there is a dielectric present as region II.

What is the ratio of the charge in region I to that in region II?

1/31/07 184 Lecture 14 26

Partially Filled Capacitor - 2Partially Filled Capacitor - 2Partially Filled Capacitor - 2Partially Filled Capacitor - 2

The voltage between the two cylinders is V, regardless of whether we are in region I or II (it is like two capacitors in parallel); since q = CV,

where the common factor is cancelled The partially filled capacitor is disconnected from the voltage

source, so that it remains charged. The dielectric is then moved further into the capacitor, so that region II now has length L-d' = 3.5 cm. Now what is the voltage difference between the cylinders?

Here the total charge is what is was before, so the new voltage is

20

ln b /a V

C q

V

20L

ln r2 / r1

1/31/07 184 Lecture 14 27

Partially Filled Capacitor - 3Partially Filled Capacitor - 3Partially Filled Capacitor - 3Partially Filled Capacitor - 3

Is a force required to hold the dielectric in place, and if so, what is its direction?• The energy of a capacitor is qV/2; since V was

reduced while q remained constant, the energy of the capacitor is lowered by inserting the dielectric into it. This means that the work was done by the capacitor+dielectric, ie, if we attached the left-hand side of the dielectric (the side outside the capacitor) to a spring the dielectric would pull on the spring.

• A force is required to keep the dielectric from moving further into the capacitor.

1/31/07184 Lecture 141 PHY 184 Spring 2007 Lecture 14.

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