tuesday, june 19, 2012 1 phys 1444 dr. andrew brandt phys 1444 lecture #5 tuesday june 19, 2012 dr....

Tuesday, June 19, 2012 1PHYS 1444 Dr. Andrew Brandt

PHYS 1444 Lecture #5

Tuesday June 19, 2012Dr. Andrew Brandt

• Short review• Chapter 24

• Capacitors and Capacitance

Tuesday, June 19, 2012 2

1 2Q QCoulomb’s Law – The Formula

A vector quantity. Newtons• Direction of electric (Coulomb) force (Newtons) is always

along the line joining the two objects.• Unit of charge is called Coulomb, C, in SI.• Elementary charge, the smallest charge, is that of an

electron: -e where

F 1Q 1 22

Q QF k

r

2r 2Q

Formula

191.602 10e C

PHYS 1444 Dr. Andrew Brandt


Vector Problems • Calculate magnitude of vectors (Ex.

force using Coulomb’s Law)• Split vectors into x and y components

and add these separately, using diagram to help determine sign

• Calculate magnitude of resultant |F|=(Fx

2+Fy2)

• Use = tan-1(Fy/Fx) to get angle



Gauss’ Law• The precise relation between flux and the enclosed charge is

given by Gauss’ Law

• 0 is the permittivity of free space in the Coulomb’s law• A few important points on Gauss’ Law

– Freedom to choose surface– Distribution of charges inside surface does not matter only total

charge – Charges outside the surface do not contribute to Qencl.

0

enclQE dA

rrÑ


Example 22-3: Spherical conductor.

A thin spherical shell of radius r0 possesses a total net charge Q that is uniformly distributed on it. Determine the electric field at points (a) outside the shell, and (b) within the shell. (c) What if the conductor were a solid sphere? *q3

Figure 22-11. Cross-sectional drawing of a thin spherical shell of radius r0 carrying a net charge Q uniformly distributed. A1 and A2 represent two gaussian surfaces we use to determine Example 22–3.Solution: a. The gaussian surface A1, outside the shell, encloses the charge Q. We know the field must be radial, so E = Q/(4πε0r2).b. The gaussian surface A2, inside the shell, encloses no charge; therefore the field must be zero.c. All the excess charge on a conductor resides on its surface, so these answers hold for a solid sphere as well.

0

enclQE dA

rrÑ

Key to these questions is how much charge is enclosed5Tuesday, June 19, 2012 PHYS 1444 Dr. Andrew Brandt

Example 22-4: Solid sphere of charge.

An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r0. Determine the electric field (a) outside the sphere (r > r0) and (b) inside the sphere (r < r0).

*q4

Solution: a. Outside the sphere, a gaussian surface encloses the total charge Q. Therefore, E = Q/(4πε0r2).b. Within the sphere, a spherical gaussian surface encloses a fraction of the charge Qr3/r0

3 (the ratio of the volumes, as the charge density is constant). Integrating and solving for the field gives E = Qr/(4πε0r0

3).


Example 22-5: Nonuniformly charged solid sphere.

Suppose the charge density of a solid sphere is given by ρE = αr2, where α is a constant. (a) Find α in terms of the total charge Q on the sphere and its radius r0. (b) Find the electric field as a function of r inside the sphere.

Solution: a. Consider the sphere to be made of a series of spherical shells, each of radius r and thickness dr. The volume of each is dV = 4πr2 dr. To find the total charge: Q = ∫ρE dV = 4παr0

5/5, giving α = 5Q/4πr05.

b. The charge enclosed in a sphere of radius r will be Qr5/r05.

Gauss’s law then gives E = Qr3/4πε0r05.



Electric Potential Energy• Concept of energy is very useful solving

mechanical problems• Conservation of energy makes solving complex

problems easier. • Defined for conservative forces (independent of

path)



• What are the differences between the electric potential and the electric field?– Electric potential (U/q)

• Simply add the potential from each of the charges to obtain the total potential from multiple charges, since potential is a scalar quantity

– Electric field (F/q)

• Need vector sums to obtain the total field from multiple charges

• Potential for a positive charge is large near a positive charge and decreases to 0 at large distances.

• Potential for the negative charge is small (large magnitude but negative) near the charge and increases with distance to 0

Properties of the Electric Potential

20

1

4

QE

r

r

0

1

4

QV

r



Capacitors (or Condensers)• What is a capacitor?

– A device that can store electric charge without letting the charge flow• What does it consist of?

– Usually consists of two oppositely charged conducting objects (plates or sheets) placed near each other without touching

– Why can’t they touch each other?• The charges will neutralize each other

• Can you give some examples?– Camera flash, surge protectors, computer keyboard, binary circuits…

• How is a capacitor different than a battery?– Battery provides potential difference by storing energy (usually chemical

energy) while the capacitor stores charge but very little energy.

11

Capacitors• A simple capacitor consists of a pair of parallel plates

of area A separated by a distance d.– A cylindrical capacitors are essentially parallel plates

wrapped around as a cylinder.

• Symbols for a capacitor and a battery:– Capacitor -||-– Battery (+) -|i- (-)

Circuit Diagram

Tuesday, June 19, 2012 PHYS 1444 Dr. Andrew Brandt


• What do you think will happen if a battery is connected (voltage is applied) to a capacitor?– The capacitor gets charged quickly, one plate positive and the other

negative with an equal amount. of charge• Each battery terminal, the wires and the plates are

conductors. What does this mean?– All conductors are at the same potential. – the full battery voltage is applied across the capacitor plates.

• So for a given capacitor, the amount of charge stored in the capacitor is proportional to the potential difference Vba between the plates. How would you write this formula?

– C is a proportionality constant, called capacitance of the device.– What is the unit?

Capacitors

baQ CV

C/V or Farad (F)

C is a property of a capacitor so does not depend on Q or V.

Normally use F or pF.


Determination of Capacitance• C can be determined analytically for capacitors w/ simple

geometry and air in between.• Let’s consider a parallel plate capacitor.

– Plates have area A each and separated by d.• d is smaller than the length, so E is uniform.

– For parallel plates E=0, where is the surface charge density.

• E and V are related• Since we take the integral from the lower potential point a to

the higher potential point b along the field line, we obtain• • So from the formula:

– What do you notice?

baV

baV

0

0ba

AQ QC

V Qd A d

C only depends on the area (A) and the separation (d) of the plates and the permittivity of the medium between them.

b aV V cos180b

aEdl b

aEdl

0

b

adl

0

b

a

Qdl

A

0

b

a

Qdl

A

0

Qb a

A

0

Qd

A

b

aE dl

rr


Example 24 – 1Capacitor calculations: (a) Calculate the capacitance of a capacitor whose plates are 20 cm x 3.0 cm and are separated by a 1.0 mm air gap. (b) What is the charge on each plate if the capacitor is connected to a 12 V battery? (c) What is the electric field between the plates? (d) Estimate the area of the plates needed to achieve a capacitance of 1F, given the same air gap.

(a) Using the formula for a parallel plate capacitor, we obtain 0 A

Cd

(b) From Q=CV, the charge on each plate is

Q

2

12 2 2 12 23

0.2 0.038.85 10 53 10 53

1 10

mC N m C N m pF

m

CV 12 2 1053 10 12 6.4 10 640C N m V C pC


Example 24 – 1(c) Using the formula for the electric field in two parallel plates

E

(d) Solving the capacitance formula for A, we obtain

V Ed E Or, since

we can obtain

0 AC

d

A

Solve for A

About 40% the area of Arlington (256km2).

V

d 4

3

121.2 10

1.0 10

VV m

m

0

0

Q

A

104 4

3 2 12 2 2

6.4 101.2 10 1.2 10

6.0 10 8.85 10

CN C V m

m C N m

0

Cd

38 2 2

12 2 2

1 1 1010 100

9 10

F mm km

C N m


Example 24 – 3Spherical capacitor: A spherical capacitor consists of two thin concentric spherical conducting shells, of radius ra and rb, as in the figure. The inner shell carries a uniformly distributed charge Q on its surface and the outer shell an equal but opposite charge –Q. Determine the capacitance of this configuration. Using Gauss’ law, the electric field outside a uniformly charged conducting sphere is 2

04

QE

r

So the potential difference between a and b isb

baa

V E dl rr

C Thus capacitance is

b

aE dr 2

04

b

a

Qdr

r 2

0 0

1

4 4

b

a

rb

ar

Q Qdr

rr

0

1 1

4 b a

Q

r r

04

a b

b a

r rQ

r r

Q

V

04a b

b a

Q

r rQ

r r

04 b a

a b

r r

r r

17PHYS 1444 Dr. Andrew Brandt

Capacitor Cont’d• A single isolated conductor can be said to have a

capacitance, C.• C can still be defined as the ratio of the charge to absolute

potential V on the conductor.– So Q=CV.

• The potential of a single conducting sphere of radius rb can be obtained as

• So its capacitance is• Although it has capacitance, a single conductor is not considered to be

a capacitor, as a second nearby charge is required to store charge

V ar

0

1 1

4 b a

Q

r r

04 b

Q

r where

0/ 4 bC Q V r

Tuesday, June 19, 2012


Capacitors in Series or Parallel• Capacitors are used in many electric circuits• What is an electric circuit?

– A closed path of conductors, usually wires connecting capacitors and other electrical devices, in which

• charges can flow• there is a voltage source such as a battery

• Capacitors can be connected in various ways.– In parallel and in Series or in combination

19

Capacitors in Parallel• Parallel arrangement provides the same

voltage across all the capacitors. – Left hand plates are at Va and right hand

plates are at Vb

– So each capacitor plate acquires charges given by the formula

• Q1=C1V, Q2=C2V, and Q3=C3V

• The total charge Q that must leave battery is thenQ=Q1+Q2+Q3=V(C1+C2+C3)

• Consider that the three capacitors behave like a single “equivalent” oneQ=CeqV= V(C1+C2+C3)

• Thus the equivalent capacitance in parallel is 1 2 3eqC C C C

For capacitors in parallel the capacitance is the sum of the individual capacitors

20PHYS 1444 Dr. Andrew Brandt

Capacitors in Series• Series arrangement is more “interesting”

When battery is connected, +Q flows to the left plate of C1 and –Q flows to the right plate of C3

This induces opposite sign charges on the other plates. Since the capacitor in the middle is originally neutral,

charges get induced to neutralize the induced charges So the charge on each capacitor is the same value, Q. (Same charge)

• Consider that the three capacitors behave like an equivalent one Q=CeqV V=Q/Ceq

• The total voltage V across the three capacitors in series must be equal to the sum of the voltages across each capacitor. V=V1+V2+V3=(Q/C1+Q/C2+Q/C3)

• Putting all these together, we obtain:• V=Q/Ceq=Q(1/C1+1/C2+1/C3)• Thus the equivalent capacitance is 1 2 3

1 1 1 1

eqC C C C

The total capacitance is smaller than the smallest C!!!

tuesday, june 19, 2012 1 phys 1444 dr. andrew brandt phys 1444 lecture #5 tuesday june 19, 2012 dr....

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