1/29/07 184 Lecture 12 1

PHY 184PHY 184

Spring 2007Lecture 12

Title: Capacitor calculations

1/29/07 184 Lecture 12 2

AnnouncementsAnnouncements

Homework Set 3 is due tomorrow morning at 8:00 am. Midterm 1 will take place in class next week on

Thursday, February 8. Practice exam will be posted in a few days. Second half of this Thursday’s lecture: review.

1/29/07 184 Lecture 12 3

Review of CapacitanceReview of Capacitance The definition of capacitance is

The unit of capacitance is the farad (F)

The capacitance of a parallel plate capacitor is given by

Variables: A is the area of each plated is the distance between the plates

C qV

1 F 1 C1 V

C 0Ad

1/29/07 184 Lecture 12 4

Consider a capacitor constructed of two collinear conducting cylinders of length L.

The inner cylinder has radius r1 andthe outer cylinder has radius r2.

Both cylinders have charge perunit length with the inner cylinderhaving positive charge and the outercylinder having negative charge.

We will assume an ideal cylindrical capacitor• The electric field points radially from the inner cylinder to the

outer cylinder.• The electric field is zero outside the collinear cylinders.

Cylindrical CapacitorCylindrical Capacitor

1/29/07 184 Lecture 12 5

Cylindrical Capacitor (2)Cylindrical Capacitor (2) We apply Gauss’ Law to get the electric field between the two

cylinder using a Gaussian surface with radius r and length L as illustrated by the red lines

… which we can rewrite to get anexpression for the electric fieldbetween the two cylinders

E

20r

rLALEA

qAdE

2 where0

0

1/29/07 184 Lecture 12 6

Cylindrical Capacitor (3)Cylindrical Capacitor (3) As we did for the parallel plate capacitor, we define the voltage

difference across the two cylinders to be V=V1 – V2.

The capacitance of a cylindrical capacitor is

C qV

L

20

ln r2 / r1

20Lln r2 / r1

1

2

0

021

ln2

22

1

2

1

rr

drr

sdEVV rr

rr

Note that C depends on geometrical factors.

1/29/07 184 Lecture 12 7

Spherical CapacitorSpherical Capacitor Consider a spherical capacitor formed by two concentric

conducting spheres with radii r1 and r2

1/29/07 184 Lecture 12 8

Spherical Capacitor (2)Spherical Capacitor (2) Let’s assume that the inner sphere has charge +q and the outer

sphere has charge –q. The electric field is perpendicular to the surface of both spheres

and points radially outward

1/29/07 184 Lecture 12 9

Spherical Capacitor (3)Spherical Capacitor (3) To calculate the electric field, we use a Gaussian surface

consisting of a concentric sphere of radius r such that r1 < r < r2

The electric field is always perpendicular to the Gaussian surface so

… which reduces toE

q40r

2…makes sense!

1/29/07 184 Lecture 12 10

Spherical Capacitor (4)Spherical Capacitor (4) To get the electric potential we follow a method similar to the one

we used for the cylindrical capacitor and integrate from the negatively charged sphere to the positively charged sphere

Using the definition of capacitance we find

The capacitance of a spherical capacitor is then

V Edrr2

r1

q

40r2 drr2

r1

q

40

1r1

1r2

C qV

q

q40

1r1

1r2

40

1r1

1r2

C 40r1r2r2 r1

1/29/07 184 Lecture 12 11

Capacitance of an Isolated SphereCapacitance of an Isolated Sphere We obtain the capacitance of a single conducting

sphere by taking our result for a spherical capacitor and moving the outer spherical conductor infinitely far away.

Using our result for a spherical capacitor…

…with r2 = and r1 = R we findC 40R

C qV

q

q40

1r1

1r2

40

1r1

1r2

…meaning V = q/40R (we already knew that!)

1/29/07 184 Lecture 12 12

ExampleExample The “plates” of a spherical capacitor have radii 38 mm

and 40 mm.

a) Calculate the capacitance. b) Calculate the area A of a parallel-plate capacitor

with the same plate separation and capacitance.

b=40 mm

a=38 mm

d

A ?

Answers: (a) 84.5 pF; (b) 191 cm2

1/29/07 184 Lecture 12 13

Clicker QuestionClicker Question Two metal objects have charges of 70pC and -

70pC, resulting in a potential difference (voltage) of 20 V between them. What is the capacitance of the system?

A) 140 pFB) 3.5 pFC) 7 pFD) 0 pF 5.3

V 20pC 70

VqC

1/29/07 184 Lecture 12 14

Clicker QuestionClicker Question Two metal objects have charges of 70pC and -70pC, resulting in

a potential difference (voltage) of 20 V between them. How does the capacitance C change if we double the charge on each object?

A) C doublesB) C is cut in halfC) C does not change

The capacitance is the constant of proportionality between change and voltage. It depends on the geometry not on the charge or voltage.

1/29/07 184 Lecture 12 15

Capacitors in CircuitsCapacitors in Circuits A circuit is a set of electrical devices connected

with conducting wires. Capacitors can be wired together in circuits in

parallel or series• Capacitors in circuits connected by wires such that

the positively charged plates are connected together and the negatively charged plates are connected together, are connected in parallel.

• Capacitors wired together such that the positively charged plate of one capacitor is connected to the negatively charged plate of the next capacitor are connected in series.

1/29/07 184 Lecture 12 16

Capacitors in ParallelCapacitors in Parallel Consider an electrical circuit with three capacitors

wired in parallel

Each of three capacitors has one plate connected to the positive terminal of a battery with voltage V and one plate connected to the negative terminal.

The potential difference V across each capacitor is the same.

We can write the charge on each capacitor as …q1 C1V q2 C2V q3 C3V

.. key point for capacitors in parallel

1/29/07 184 Lecture 12 17

Capacitors in Parallel (2)Capacitors in Parallel (2) We can consider the three capacitors as one equivalent

capacitor Ceq that holds a total charge q given by

We can now define Ceq by

A general result for n capacitors in parallel is If we can identify capacitors in a circuit that are wired

in parallel, we can replace them with an equivalent capacitance

q q1 q2 q3 C1V C2V C3V C1 C2 C3 V

Ceq C1 C2 C3Ceq Ci

i1

n

q CeqV

1/29/07 184 Lecture 12 18

Capacitors in SeriesCapacitors in Series Consider a circuit with three capacitors wired in series The positively charged plate of C1 is connected to the

positive terminal of the battery. The negatively charge plate of C1 is connected to the

positively charged plate of C2.

The negatively charged plate of C2 is connected to thepositively charge plate of C3.

The negatively charge plate of C3 is connected to thenegative terminal of the battery.

The battery produces an equal charge q on each capacitor because the battery induces a positive charge on the positive place of C1, which induces a negative charge on the opposite plate of C1, which induces a positive charge on C2, etc.

.. key point for capacitors in series

1/29/07 184 Lecture 12 19

Capacitors in Series (2)Capacitors in Series (2) Knowing that the charge is the same on all three capacitors

we can write

We can express an equivalent capacitance Ceq as

We can generalize to n capacitors in series

If we can identify capacitors in a circuit that are wired in series, we can replace them with an equivalent capacitance.

V V1 V2 V3 qC1

qC2

qC3

q1C1

1C2

1C3

V qCeq

1Ceq

1C1

1C2

1C3

1Ceq

1Cii1

n

1/29/07 184 Lecture 12 20

Clicker QuestionClicker Question C1=C2=C3=30 pF are placed in series. A battery supplies 9 V. What

is the charge q on each capacitor?

A) q=90 pCB) q=1 pCC) q=3 pCD) q=180 pC

Answer: 90 pC

Ceq = 10 pF

1/29/07 184 Lecture 12 21

Clicker QuestionClicker Question C1=C2=C3=1 pF are placed in parallel. What is the voltage of the battery if the total charge of the capacitor

arrangement, q1+q2+q3, is 90 pC?

A) 180 VB) 10 VC) 9 VD) 30 V

Answer: 30 volts

Ceq = 3 pF