Physics 169 Luis Anchordoqui Kitt Peak National Observatory Wednesday, January 30, 19 1 SECTION 1 Electric Charge

Physics 169 Luis anchordoqui Kitt Peak National Observatory Wednesday, January 30, 19 1 SECTION 1 Electric Charge

SECTION OBJECTIVES

■ Understand the basic properties of electric charge. PROPERTIES OF ELECTRIC CHARGE ■ Differentiate between con- You have probably noticed that after running a plastic comb through your ductors and insulators. hair on a dry day, the comb attracts strands of your hair or small pieces of ■ Distinguish between charging paper. A simple experiment you might try is to rub an inflated balloon back by contact, charging by induction, and charging and forth across your hair. You may find that the balloon is attracted to your by polarization. hair, as shown in Figure 1(a). On a dry day, a rubbed balloon will stick to the wall of a room, often for hours. When materials behave this way, they are said to be electrically charged. Experiments such as these work best on a dry day because excessive moisture can provide a pathway for charge to leak off a charged object. You can give your body an electric charge by vigorously rubbing your shoes on a wool rug or by sliding across a car seat. You can then remove the charge on your body by lightly touching another person. Under the right conditions, Table 1 Conventions you will see a spark just before you touch, and both of you will feel a slight tingle. for Representing Charges Another way to observe static electricity is to rub two balloons across your and Electric Field Vectors hair and then hold them near one another, as shown in Figure 1(b). In this case, you will see the two balloons pushing each other apart. Why is a rubbed + Positive charge +q balloon attracted to your hair but repelled by another rubbed balloon? − Negative charge There are two kinds of electric charge −q The two balloons must have the same kind of charge because each became Electric field vector E charged in the same way. Because the two charged balloons repel one another, we see that like charges repel. Conversely, a rubbed balloon and your hair, Electric field lines which do not have the same kind of charge, are attracted to one another. Thus, unlike charges attract. 1.1 Electric Charge

(a) (b)

Figure 1 (a) If you rub a balloon across your hair on a dry day, the balloon and your hair become charged and attract each other. (b) Tw o charged balloons, on the other hand, repel each other. (a) If you rub a balloon across your hair on a dry day 558 Chapter 16 the balloon and your hair become charged and attract each other

(b) Two charged balloons, on the other hand, repel each other. The two balloons must have the same kind of charge because each became charged in the same way Because two charged balloons repel one another we see that like charges repel Conversely ☛ a rubbed balloon and your hair which do not have the same kind of charge are attracted to one another ☛ unlike charges attract Wednesday, January 30, 19 2 708 CHAPTER 23 • Electric Fields

Rubber Rubber

– – – – – – – – – – F F Glass Rubber + + – – + + + + – – – – + – F (a) (b) Figure 23.1 (a) A negatively charged rubber rod suspended by a thread is attracted to a positively charged glass rod. (b) A negatively charged rubber rod is repelled by another negatively charged rubber rod.

have a positive charge, and any charged object repelled by a charged rubber rod (or attracted to a charged glass rod) must have a negative charge. Attractive electric forces are responsible for the behavior of a wide variety of com- mercial products. For example, the plastic in many contact lenses, etaﬁlcon, is made up of molecules that electrically attract the protein molecules in human tears. These protein molecules are absorbed and held by the plastic so that the lens ends up being primarily composed of the wearer’s tears. Because of this, the lens does not behave as a foreign object to the wearer’s eye, and it can be worn comfortably. Many cosmetics also take advantage of electric forces by incorporating materials that are electrically attracted to skin or hair, causing the pigments or other chemicals to stay put once they are applied. Another important aspect of electricity that arises from experimental observa- Electric charge is conserved tions is that electric charge is always conserved in an isolated system. That is, when one object is rubbed against another, charge is not created in the process. The Charge is electriﬁconserveded state and is due quantized to a transfer of charge from one object to the other. One object gains some amount of negative charge while the other gains an equal amount Whenof apositive glass charge.rod is Forrubbed example, with when silk a glass rod is rubbed with silk, as in Figure – 23.2, the silk obtains a negative charge that is equal in magnitude to the posi- – + electrons are transferred from the glass to the silk ++ tive charge on the glass rod. We now know from our understanding of atomic struc- – ++ Because of conservation of charge + – ture that electrons are transferred from the glass to the silk in the rubbing process. – each electron adds negative charge to the silk – Similarly, when rubber is rubbed with fur, electrons are transferred from the fur to and thean equalrubber ,positive giving the charge rubber is a leftnet negativebehind chargeon the and rod the fur a net positive Also charge.☛ because This process charges is consistentare transferred with the factin discrete that neutral, bundles uncharged matter contains as many positive charges (protons within atomic nuclei) as negative chargescharges on (electrons).the two objects are e, 2 e, 3 e, In 1909, Robert Millikan (1868–1953)± ± discovered± that··· electric charge always Figure 23.2 When a glass rod is occurs as some integral multiple of a fundamental amount of charge e (see Section 708 CHAPTER 23 • Electric Fields rubbed with silk, electrons are transferred from the glass to the 25.7). In modern terms, the electric charge q is said to be quantized, where q is the silk. Because of conservation of standard symbol used leftfor charge as a variable. That is, electric charge exists as charge, each electron adds nega- discrete “packets,” and we can write q ϭ Ne, where N is some integer. Other experi- tive charge to the silk, and an equal ments in the same periodA negatively showed that chargedthe electron rubber has a charge rod Ϫ e and the proton positive charge is left behind on has a charge of equalsuspended magnitude butby aopposite thread sign is attractedϩ e. Some particles, such as the rod. Also, because the charges are transferred in discrete bundles, the neutron,Rubber have no charge.to a positively charged glass rod the charges on the twoRubber objects are From our discussion thus far, we conclude that electric charge has the following im- Ϯ e, or Ϯ 2e, or Ϯ 3e, and so on. Fportant properties: – – – – – right – – – – – F F Glass Rubber + + – – A negatively charged rubber rod + + + + – – – – + – F is repelled by another negatively charged (a) (b) rubber rod Figure 23.1 (a) A negatively charged rubber rod suspended by a thread is attracted to a positively charged glass rod. (b) A negatively charged rubber rod is repelled by another negatively charged rubber rod. Wednesday, January 30, 19 3

have a positive charge, and any charged object repelled by a charged rubber rod (or attracted to a charged glass rod) must have a negative charge. Attractive electric forces are responsible for the behavior of a wide variety of com- mercial products. For example, the plastic in many contact lenses, etaﬁlcon, is made up of molecules that electrically attract the protein molecules in human tears. These protein molecules are absorbed and held by the plastic so that the lens ends up being primarily composed of the wearer’s tears. Because of this, the lens does not behave as a foreign object to the wearer’s eye, and it can be worn comfortably. Many cosmetics also take advantage of electric forces by incorporating materials that are electrically attracted to skin or hair, causing the pigments or other chemicals to stay put once they are applied. Another important aspect of electricity that arises from experimental observa- Electric charge is conserved tions is that electric charge is always conserved in an isolated system. That is, when one object is rubbed against another, charge is not created in the process. The electriﬁed state is due to a transfer of charge from one object to the other. One object gains some amount of negative charge while the other gains an equal amount of positive charge. For example, when a glass rod is rubbed with silk, as in Figure – 23.2, the silk obtains a negative charge that is equal in magnitude to the posi- – + ++ tive charge on the glass rod. We now know from our understanding of atomic struc- – ++ + – ture that electrons are transferred from the glass to the silk in the rubbing process. – – Similarly, when rubber is rubbed with fur, electrons are transferred from the fur to the rubber, giving the rubber a net negative charge and the fur a net positive charge. This process is consistent with the fact that neutral, uncharged matter contains as many positive charges (protons within atomic nuclei) as negative charges (electrons). In 1909, Robert Millikan (1868–1953) discovered that electric charge always Figure 23.2 When a glass rod is occurs as some integral multiple of a fundamental amount of charge e (see Section rubbed with silk, electrons are transferred from the glass to the 25.7). In modern terms, the electric charge q is said to be quantized, where q is the silk. Because of conservation of standard symbol used for charge as a variable. That is, electric charge exists as charge, each electron adds nega- discrete “packets,” and we can write q ϭ Ne, where N is some integer. Other experi- tive charge to the silk, and an equal ments in the same period showed that the electron has a charge Ϫ e and the proton positive charge is left behind on has a charge of equal magnitude but opposite sign ϩ e. Some particles, such as the rod. Also, because the charges are transferred in discrete bundles, the neutron, have no charge. the charges on the two objects are From our discussion thus far, we conclude that electric charge has the following im- Ϯ e, or Ϯ 2e, or Ϯ 3e, and so on. portant properties: Chapter 2

Electric Force & Electric Field

2.1 Electric Force

1.2 Electric Force ElectricThe electric force forcebetween between two twochargescharges q 1 and q 2 describedq1 and byq 2☛ Coulomb’scan be described Law by Coulomb’s Law.

F~12 = Force on q 1 exerted by q2 ~ 1 q1q2 F12 = 2 rˆ12 F~12 = Force on q14exerted⇡✏0 · byr12q2 · ~r 12 rˆ = ☛ unit vector which locates particle 1 relative to particle 2 12 ~r ~ 1 q1q2 12 F12 = 4º≤ r2 rˆ12 | | i.e. ~r = ~0r · 12~r · 12 1 2 ~r12 wherer ˆ12 = is the unit vector which locates particle 1 relative to particle 2. • q 1 ,q 2 are ~relectrical charges in units of Coulomb (C) | 12| 19 • Charge is quantized ☛ electron carries 1.602 10 C i.e. ~r12 = ~r1 ~r2 ° 12⇥ 2 2 • Permittivity of free space ✏0 =8.85 10 C /Nm q1, q2 are electrical charges in units of Coulomb⇥ (C) • Charge is quantized Wednesday,• January 30, 19 19 4 Recall 1 electron carries 1.602 10° C £ 12 2 2 ≤ = Permittivity of free space = 8.85 10° C /N m • 0 £ COULOMB’S LAW:

(1) q1, q2 can be either positive or negative. COULOMB’S LAW:

(1) q 1 ,q 2 can be either positive or negative

(2) If q 1 ,q 2 are of same sign force experienced by q 2 is in direction away from q 1 i.e. ☛ repulsive

(3) Force on q 2 exerted by q 1 :

~ 1 q2q1 F21 = 2 rˆ21 4⇡✏0 · r21 · BUT r12 = r21 = distance between q1,q2

~r 21 ~r 2 ~r 1 ~r 12 rˆ21 = = = = rˆ12 r21 r21 r12

F~ = F~ Newton’s 3rd Law 21 12 Wednesday, January 30, 19 5 2.2. THE ELECTRIC FIELD 9

(2) If q1, q2 are of same sign, then the force experienced by q1 is in direction away from q2, that is, repulsive.

(3) Force on q2 exerted by q1:

~ 1 q2q1 F21 = 2 rˆ21 4º≤0 · r21 · BUT:

r12 = r21 = distance between q1,q2 ~r21 ~r2 ~r1 ~r12 rˆ21 = = ° = ° = rˆ12 r21 r21 r12 °

F~ = F~ Newton’s 3rd Law ) 21 ° 12

SYSTEM WITH MANY CHARGES: SYSTEM WITH MANY CHARGES:

The total force experienced by charge q1 is the vector sum of the forces on q1 exerted by other charges.

Total force experienced by charge q 1 vector sum of forces on q 1 exerted by other charges

F~1 = Force experienced by q1 F~1 = Force experienced by q1 = F~1,2 + F~1,3 + F~1,4 + + F~1,N = F~ + F~ + F~ + + F~··· 1,2 1,3 1,4 ··· 1,N PRINCIPLE OF SUPERPOSITION: N N PRINCIPLE OF SUPERPOSITION F~ = F~ F~1 = F~1,j 1 1,j j=2 jX=2 X Wednesday, January 30, 19 6 2.2 The Electric Field

While we need two charges to quantify the electric force, we define the electric field for any single charge distribution to describe its eÆect on other charges. 1.3 Electric Field While we need two charges to quantify electric force we define electric field for any single charge distribution 2.2. THE ELECTRIC FIELD 10 to describe its effect on other charges

Total fo rce F~ = F~ + F~ + + F~ 1 2 ··· N Total force F~ = F~ + F~ + + F~ 1 2 ··· N The electric ﬁeld is deﬁned as

Electric field is deﬁned asF~ lim = E~ q0 0 q F~ ! 0 E~ =lim q 0 0! q0

(a) E-ﬁeld due to a single charge qi:

Wednesday, January 30, 19 7

From the deﬁnitions of Coulomb’s Law, the force experienced at location of q0 (point P)

~ 1 q0qi F0,i = 2 rˆ0,i 4º≤0 · r0,i ·

wherer ˆ0,i is the unit vector along the direction from charge qi to q0,

rˆ0,i = Unit vector from charge qi to point P

=ˆri (radical unit vector from qi) F~ Recall E~ = lim q0 0 ! q0 ) E-ﬁeld due to qi at point P: ~ 1 qi Ei = 2 rˆi 4º≤0 · ri ·

where ~ri = Vector pointing from qi to point P, thusr ˆi = Unit vector pointing from qi to point P Note: (1) E-ﬁeld is a vector.

(2) Direction of E-ﬁeld depends on both position of P and sign of qi. (b) E-ﬁeld due to system of charges:

Principle of Superposition: In a system with N charges, the total E-ﬁeld due to all charges is the vector sum of E-ﬁeld due to individual charges. 2.2. THE ELECTRIC FIELD 10

Total force F~ = F~ + F~ + + F~ 1 2 ··· N The electric ﬁeld is deﬁned as

F~ lim = E~ q0 0 ! q0

(a) E-field due to a single charge qi: (i) E-field due to a single charge qi From definitions of Coulomb’s Law

Fromforce the definitionsexperienced of Coulomb’s at location Law of, q the 0 (point P ) force experienced~ at location1 of q0 (pointq0qi P) F0,i = 2 rˆ0,i 4⇡✏0 · r0,i · rˆ 0 ,i ☛ unit vector along direction from charge q i to q0 ~ 1 q0qi F~F0,i = 2 rˆ0,i Recall E~ =lim )4º≤0E~· -fieldr0,i · due to q i at point P q0 0 q wherer ˆ0,i is the unit! vector0 along the direction from charge qi to q0, ~ 1 qi rˆ0,i = Unit vector fromEi = charge qi to point2 P rî 4⇡✏0 · ri · =ˆri (radical unit vector from qi) ~r i ☛ vector pointing from q i to point P F~ Recall E~ = lim rˆ i ☛ unit vector pointing from q i to pointP q0 0 ! q0 ) E-field due to qi at point P: Note: ~ ~ 1 qi (1) E -field is a vectorEi = 2 rî 4º≤0 · ri · (2) Direction of E ~ -field depends on both position of P and sign of qi where ~ri = Vector pointing from qi to point P, thusr î = Unit vector pointing from qi to point P Note:Wednesday, January 30, 19 8 (1) E-field is a vector.

(2) Direction of E-ﬁeld depends on both position of P and sign of qi. (b) E-ﬁeld due to system of charges:

Principle of Superposition: In a system with N charges, the total E-field due to all charges is the vector sum of E-field due to individual charges. 2.2. THE ELECTRIC FIELD 11 (ii) E~ -field due to system of charges: Principle of Superposition1 q i.e. ~ ~ i E = Ei = 2 rî ~ In a system with N charges4º≤0 ☛ totalri E -field due to all charges Xi Xi vector sum of E ~ -field due to individual charges (c) Electric Dipole 1 qi i.e. ☛ E~ = E~ = rˆ i 4⇡✏ r2 i i 0 i i (iii) Electric Dipole X X

System of equal and opposite charges separated by a distance d. System ofFigure equal 2.1: and Anopposite electric charges dipole. separated (Direction by ofa distance d d~ from negative to positive charge) Electric Dipole Moment ☛ p~ = qd~ = qddˆ

p = qd Electric Dipole Moment Wednesday, January 30, 19 9 p~ = qd~ = qddˆ p = qd

Example: E~ due to dipole along x-axis

Consider point P at distance x along the perpendicular axis of the dipole p~ :

E~ = E~+ + E~ ° (E-ﬁeld"" (E-ﬁeld due to +q) due to q) °

Notice: Horizontal E-ﬁeld components of E~+ and E~ cancel out. °

) Net E-ﬁeld points along the axis opposite to the dipole moment vector. 2.2. THE ELECTRIC FIELD 11

~ ~ 1 qi i.e. E = Ei = 2 rˆi 4º≤0 ri Xi Xi

System of equal and opposite charges separated by a distance d. Figure 2.1: An electric dipole. (Direction of d~ from negative to positive charge)

Electric Dipole Moment p~ = qd~ = qddˆ p = qd

Example: E~ due to dipole along x-axis Example: E~ due to dipole along x -axis

Consider point P at distance x along perpendicular axis of dipole p~ Consider point P at distance x along the perpendicular axis of the dipole p~ : ~ ~ ~ E = E~ =E+ E~+ + + EE~ ° (E-field"" (E-field dueE-field to +dueq) to+q dueE to-fieldq )due to q °

Notice:Wednesday, Horizontal January 30, E-ﬁeld 19 components of E~+ and E~ cancel out. 10 °

) Net E-ﬁeld points along the axis opposite to the dipole moment vector. 2.2. THE ELECTRIC FIELD 11

~ ~ 1 qi i.e. E = Ei = 2 rˆi 4º≤0 ri Xi Xi

System of equal and opposite charges separated by a distance d. Figure 2.1: An electric dipole. (Direction of d~ from negative to positive charge)

Electric Dipole Moment p~ = qd~ = qddˆ p = qd

Example: E~ due to dipole along x-axis

Consider point P at distance x along the perpendicular axis of the dipole p~ :

E~ = E~+ + E~ ° (E-ﬁeld"" (E-ﬁeld due to +q) due to q) °

Notice: Horizontal E-ﬁeld components of E~+ and E~ cancel out. ~ ° Notice: Horizontal E ~ -ﬁeld components of E~ + and E cancel out

) Net E-field points along the axis opposite to the dipole moment vector. 2.3. CONTINUOUS CHARGE DISTRIBUTION 12 ) Net E ~ points along axis parallel but opposite to dipole moment vector Magnitude of E-field~ = 2E+ cos µ Magnitude of E -field =2E+ cos ✓ 1 q ) E =2 2 cos ✓ 4⇡✏0 · r ! E+ or E magnitude! E+ or E °magnitude But 1 q E =2d |2 {z } cos µ r =) + x2 2 2 µ z4º≤}|0 · r { ∂ r⇣ ⌘ d/2 d 2 cos ✓ = But r = + x2 r s 2 1 ≥ p¥ E = d/2 ) d 2 3 4⇡✏0 cos· µ 2= 2 [x + r2 ] (p = qd) 1 p Wednesday, January 30, 19 E = 11 ) d 3 4º≤0 · 2 2 2 [x +(2 ) ] (p = qd) Special case: When x d ¿

2 d 2 3 3 d 2 3 [x +( ) ] 2 = x [1 + ( ) ] 2 2 2x Binomial Approximation: • (1 + y)n 1+ny if y 1 º ø 1 p 1 E-ﬁeld of dipole + 3 3 4º≤0 · x ª x

1 Compare with E-ﬁeld for single charge • r2 Result also valid for point P along any axis with respect to dipole • 2.3 Continuous Charge Distribution

E-ﬁeld at point P due to dq:

~ 1 dq dE = 2 rˆ 4º≤0 · r · Special case ☛ When x d d 2 3 d 2 3 x2 + 2 = x3 1+ 2 2 2x ⇥ ⇤ ⇥ ⇤ • Binomial Approximation

n (1 + y) 1+ny if y 1 ⇡ ⌧

~ 1 p 1 E ﬁeld of dipole 3 3 ' 4⇡✏0 · x / x 1 • Compare with E ~ -ﬁeld for single charge r2

• Result also valid for point P along any axis with respect to dipole

Wednesday, January 30, 19 12 2.3. CONTINUOUS CHARGE DISTRIBUTION 12

Magnitude of E-ﬁeld = 2E+ cos µ

E+ or E magnitude! ° 1 q ) E =2 2 cos µ µ z4º≤}|0 · r { ∂

d 2 But r = + x2 s 2 ≥ ¥ d/2 cos µ = r 1 p E = ) d 3 4º≤0 · 2 2 2 [x +(2 ) ] (p = qd) Special case: When x d ¿

2 d 2 3 3 d 2 3 [x +( ) ] 2 = x [1 + ( ) ] 2 2 2x Binomial Approximation: • (1 + y)n 1+ny if y 1 º ø 1 p 1 E-ﬁeld of dipole + 3 3 4º≤0 · x ª x

1 Compare with E-ﬁeld for single charge • r2 Result also valid for point P along any axis with respect to dipole • 2.3 Continuous Charge Distribution 1.4 Continuous Charge Distribution

E -field at point P due to dq E-field~ at point1 P duedq to dq: dE = 2 rˆ 4⇡✏0 · r · ~ 1 dq dE = 2 rˆ ) E-field due to charge4º≤ distribution0 · r · 1 dq E~ = dE~ = rˆ 4⇡✏ · r2 · Z Z 0 (1) Take advantage of symmetry of system to simplify integral

(2) To write down small charge element dq ☛ 1 D dq = ds = linear charge density ds = small length element 2 D dq = dA = surface charge density dA = small area element 3 D dq = ⇢dV ⇢ = volume charge density dV = small volume element Wednesday, January 30, 19 13 2.3. CONTINUOUS CHARGE DISTRIBUTION 13

) E-ﬁeld due to charge distribution: 1 dq E~ = dE~ = rˆ 2 ˆ ˆ 4º≤0 · r · Volume Volume (1) In many cases, we can take advantage of the symmetry of the system to simplify the integral. (2) To write down the small charge element dq: 1-D dq = ∏ ds ∏ = linear charge density ds = small length element 2-D dq = æ dA æ = surface charge density dA = small area element 3-D dq = Ω dV Ω = volume charge density dV = small volume element

Example 1:Example Uniform 1 lineUniform of charge line of charge

charge per unit length charge= ∏ per unit length =

(1) Symmetry considered ☛ -field from + z and z directions cancel along (1) Symmetry considered: The E-field fromE +z and z directions cancel along ° z-directionz-direction, ) Only, horizontal ) Only E-fieldhorizontal components E -field need components to be considered. need to be considered (2) For each element of length dz, charge dq = ∏dz ) Horizontal(2) For each E-field element at point P of due length to element dz dz, charge= dq = dz ~ 1 ∏dz dE cos µ = 2 cos µ ) Horizontal E -field at point P |due| to element4º≤0 · r dz is 1 dz dEdz dz = dE~ cos ✓ = |cos{z✓ } E-field due to entire line2 charge at point P | | ) 4⇡✏0 · r dE dzL/2 Wednesday, January 30, 19 1 ∏dz 14 E = 2 cos µ | {zˆ 4º≤}0 · r L/2 ° L/2 ∏ dz =2 2 cos µ ˆ 4º≤0 · r 0 2.3. CONTINUOUS CHARGE DISTRIBUTION 13

E-ﬁeld due to charge distribution: 1 dq E⇤ = dE⇤ = rˆ 2 ˆ ˆ 4⇥0 · r · Volume Volume (1) In many cases, we can take advantage of the symmetry of the system to simplify the integral. (2) To write down the small charge element dq: 1-D dq = ⇤ ds ⇤ = linear charge density ds = small length element 2-D dq = ⌃ dA ⌃ = surface charge density dA = small area element 3-D dq = ⇧ dV ⇧ = volume charge density dV = small volume element Example 1: Uniform line of charge

charge per unit length = ⇤

(1) Symmetry considered: The E-field from +z and z directions cancel along z-direction, Only horizontal E-field components need to be considered. (2) For each element of length dz, charge dq = ⇤dz Horizontal E-field at point P due to element dz = ⌥ 1 ⇤dz dE cos ⇥ = 2 cos ⇥ | | 4⌅0 · r

dEdz ) E -ﬁeld due to entire line charge at point P ⇤ ⇥ ⌅ E-ﬁeld due to entire line charge at point P L/2 1 dz E = L/2 2 cos ✓ L/2 4⇡✏0 · r Z 1 ⇤dz E = 2 cos ⇥ L/2 ˆdz 4⌅0 · r =2 L/2 cos ✓ 4⇡✏ · r2 Z0 0 L/2 ⇤ dz =2 2 cos ⇥ ˆ 4⌅0 · r 0 To calculate this integral ☛

• First, notice that x is ﬁxed, but z,r, ✓ all varies

• Change of variable (from z to ✓ )

Wednesday, January 30, 19 15 (1) 2 z = x tan ✓ ) dz = x sec ✓d✓ 2 2 2 x = r cos ✓ ) r = x sec ✓ z =0 ✓ =0 L/2 (2) When z = L/2 ✓ = ✓ where tan ✓ = 0 0 x ✓0 x sec2 ✓d✓ E =2 cos ✓ · 4⇡✏ x2 sec2 ✓ · 0 Z0 ✓0 1 =2 cos ✓d✓ · 4⇡✏ x · 0 Z0 1 ✓0 =2 (sin ✓) · 4⇡✏0 · x · 0 1 =2 sin ✓0 · 4⇡✏0 · x · 1 L/2 =2 · 4⇡✏0 · x · 2 2 L x + 2 r Wednesday, January 30, 19 ⇣ ⌘ 16 1 L E = along x-direction 4⇡✏0 · 2 2 L x x + 2 r ⇣ ⌘ Important limiting cases 1 L (1) x L : E + 2 4⇡✏0 · x But L = Total charge on rod ) System behave like a point charge 1 L (2) L x : E + L 4⇡✏0 · x · 2

Ex = 2⇡✏0x

ELECTRIC FIELD DUE TO INFINITELY LONG LINE OF CHARGE

Wednesday, January 30, 19 17 2.3. CONTINUOUS CHARGE DISTRIBUTION 15

Example 2: Ring of Charge Example 2 Ring of Charge

E-ﬁeld E-ﬁeldat a height at a height z abovez above a aring ring of of charge of radius R charge of radius R

(1) Symmetry(1) Symmetry considered: considered For every charge ☛ elementFor everydq considered, charge there element exists dq considered, dq where the horizontal E⌃ field components cancel. ⇤ Overall E-field lies along z-direction. ~ ⇥there exists dq 0 where horizontal E field components cancel (2) For each element of length dz, charge Wednesday, January 30, 19 18

dq = ⇤ ds · Linear Circular charge density length element

dq = ⇤ Rd⇧, where ⇧ is the angle measured· on the ring plane

Net E-ﬁeld along z-axis due to dq: 1 dq dE = 2 cos ⇥ 4⌅0 · r · 2.3. CONTINUOUS CHARGE DISTRIBUTION 15

Example 2: Ring of Charge

E-ﬁeld at a height z above a ring of charge of radius R

(1) Symmetry considered: For every charge element dq considered, there exists dq where the horizontal E⌃ ﬁeld components cancel. ⇤ Overall E-ﬁeld lies along z-direction. (2)⇥ For each element of length dz , charge (2) For each element of length dz, charge dq = ds · dq = ⇤ ds · LinearLinear Circular Circular chargecharge density densitylength length element element

dqdq= ⇤=Rd ⇧Rd, where ⇧ is the angle measured· on· the ring plane where is angle measured on ring plane

Net) E-ﬁeldNet E along -ﬁeldz-axis along due z to-axisdq: due to dq 1 dq dE = 2 cos ⇥ 4⌅10 · r dq· dE = 2 cos ✓ 4⇡✏0 · r ·

Wednesday, January 30, 19 19 Total E -ﬁeld = dE Z 2⇡ 1 R d z = cos ✓ (cos ✓ = ) 4⇡✏ · r2 · r Z0 0 Note: Here in this case, ✓ ,R and r are ﬁxed as varies! BUT we want to convert r, ✓ to R, z 1 Rz 2⇡ E = d 4⇡✏ · r3 0 Z0 1 (2⇡R)z z E = 2 2 3/2 along -axis 4⇡✏0 · (z + R )

BUT ☛ (2⇡R)= total charge on ring

Wednesday, January 30, 19 20 2.3. CONTINUOUS CHARGE DISTRIBUTION 16

Total E-ﬁeld = dE ˆ 2 1 ⇤Rd⌃ z = 2 cos ⇥ (cos ⇥ = ) ˆ0 4⌅0 · r · r

Note: Here in this case, ⇥,R and r are ﬁxed as ⌃ varies! BUT we want to convert r, ⇥ to R, z. 1 ⇤Rz 2 E = 3 d⌃ 4⌅0 · r ˆ0

1 ⇤(2⌅R)z E = 2 2 3/2 along z-axis 4⌅0 · (z + R )

BUT: ⇤(2⌅R) = total charge on the ring

Example 3:Example E-ﬁeld from 3 a disk E-field of surface chargefrom density a disk⇧ of surface charge density

We ﬁnd E-ﬁeld of a disk by integrating concentric rings of charges 2.3. CONTINUOUS CHARGE DISTRIBUTION 17

We ﬁnd the E-ﬁeld of a disk by view from top Total chargeintegrating of ring concentric rings of dqcharges.= (2⇡r dr) Total charge of ring · Area of ring dq = ⇤ (2⇥rdr) · Area of the ring| {z } ⌥ ⌦

Recall from Example 2: 1 dq z E-ﬁeld from ring: dE = 2 2 3/2 4⇥0 · (z + r )

Wednesday, January 30, 19 1 R 2⇥⇤rdr z 21 E = 2 2 ·3/2 4⇥0 ˆ0 (z + r ) 1 R rdr = 2⇥⇤z 2 2 3/2 4⇥0 ˆ0 (z + r ) Change of variable: • u = z2 + r2 (z2 + r2)3/2 = u3/2 ⇥ du =2rdr rdr= 1 du ⇥ ⇥ 2 Change of integration limit: • r =0 ,u= z2 2 2 ⌃ r = R,u= z + R

z2+R2 1 1 3/2 E = 2⇥⇤z u du 4⇥0 · ˆz2 2

1/2 3/2 u 1/2 BUT: u du = = 2u ˆ 1/2 2 2 1 1/2 z +R E = ⇤z ( u ) 2 z2 0 1 1 1 = ⇤z + 2 2 20 ⇥⇤z + R z ⇤

⇤ z E = 1 2 2 20 ⌅ ⇤z + R ⇧ Recall from Example 2 1 dq z E-ﬁeld from ring ☛ dE = 2 2 3/2 4⇡✏0 · (z + r ) 1 R 2⇡r dr z E = · ) 4⇡✏ (z2 + r2)3/2 0 Z0 1 R rdr = 2⇡z 4⇡✏ (z2 + r2)3/2 0 Z0 • Change of variable: u = z2 + r2 (z2 + r2)3/2 = u3/2 ) 1 du =2rdr rdr = du ) ) 2

Wednesday, January 30, 19 22 • Change of integration limit: r =0 u = z2 r = Ru= z2 + R2 ⇢ z2+R2 1 1 3/2 ) E = 2⇡z u du 4⇡✏0 · z2 2 BUT Z 1/2 3/2 u 1/2 u du = = 2u 1/2 Z z2+R2 1 1/2 ) E = z ( u ) 2✏0 z2 1 1 1 = z + 2✏0 pz2 + R2 z !

z E = 1 2✏0 " pz2 + R2 #

Wednesday, January 30, 19 23 2.4. ELECTRIC FIELD LINES 18

VERY IMPORTANT LIMITING CASE: VERY IMPORTANT LIMITING CASE If R z, that is if we have an inﬁnite sheet of charge with charge density ⇥:⇤ If R z , that is if we have an inﬁnite sheet of charge with charge ⇥ z density E = 1 z2 2 E = 210 ⇤ ⇧z + R ⌅ 2✏ ⇥ p z2 2 0 " 1 z + R # ⌅ 20 R ⇥ z 1 ' 2✏0 " R# E ⇡ 2✏ ⇥ E 0 ⇥ 20

E-field is normal to the charged surface Figure 2.2: E-field due to an infi- E-field is normal to charged surface nite sheet of charge, charge density = ⇥

Wednesday, January 30, 19 24

Q: What’s the E-ﬁeld belows the charged sheet?

2.4 Electric Field Lines

To visualize the electric ﬁeld, we can use a graphical tool called the electric ﬁeld lines.

Conventions: 1. The start on position charges and end on negative charges. 2. Direction of E-ﬁeld at any point is given by tangent of E-ﬁeld line.

3. Magnitude of E-field at any point is proportional to number of E-field lines per unit area perpendicular to the lines. 1.5 Electric Field Lines To visualize electric field we can use a graphical tool called electric field lines

Conventions

1. Start on positive charges and end on negative charges

2. Direction of E-ﬁeld at any point is given by tangent of E-ﬁeld line

3. Magnitude of E-ﬁeld at any point proportional to number of E-ﬁeld lines per unit area perpendicular to lines

Wednesday, January 30, 19 25 2.4. ELECTRIC FIELD LINES 19

Uniform E-ﬁeld Non-uniform E-ﬁeld

Wednesday, January 30, 19 26 2.4. ELECTRIC FIELD LINES 19

~ ~ ~ +q EP1 > EP2 E = 2 rˆ 4⇡✏0r

Wednesday, January 30, 19 27 2.4. ELECTRIC FIELD LINES 19

Inﬁnite sheet of charge

E = 2✏0

Wednesday, January 30, 19 28 2.4. ELECTRIC FIELD LINES 20

Wednesday, January 30, 19 29 2.4. ELECTRIC FIELD LINES 20

E~at point O =0

Wednesday, January 30, 19 30 2.5. POINT CHARGE IN E-FIELD 21

2.5 Point Charge in E-field 1.6 PointWhen weCharge place a charge inq in E-field an E-field E , the force experienced by the charge is When we place a charge q in an E F-field= qE =E ~ m,forcea experienced by charge is ~ ~ Applications: Ink-jetF printer,= qE TV= cathoderaym~a tube. ☛ ApplicationsExample : Ink-jet printer, TV cathode ray tube Example Ink particle has mass m, charge q (q<0 here) Ink particle has mass m & charge q (q< 0 here)

Assume Assumethat mass that mass of ofinkdrop inkdrop is small,small, what’s what’s the deﬂectiondeﬂectiony of theof charge?charge? Solution: Wednesday, January 30, 19 31 First, the charge carried by the inkdrop is negtive, i.e. q<0.

Note: qE points in opposite direction of E .

Horizontal motion: Net force = 0

L = vt (2.1) 2.5. POINT CHARGE IN E-FIELD 21

2.5 Point Charge in E-ﬁeld

When we place a charge q in an E-ﬁeld E , the force experienced by the charge is

F = qE = ma

Applications: Ink-jet printer, TV cathoderay tube.

Example:

Ink particle has mass m, charge q (q<0 here)

Assume that mass of inkdrop is small, what’s the deﬂection y of the charge?

Solution:

First, the charge carried by the inkdrop is negtive, i.e. q<0.

Solution ☛ Charge carried by inkdrop is negative ☛ q<0

Note: Note:q E ~ points in qoppositeE points direction in oppositeof E~ direction of E .

Horizontal motion ☛ Net force =0 ) L = vt Vertical motion ☛ qE~ m~g ,qis negative Horizontal motion: Net force = 0 | || | ) Net force = q E = ma ☛ Newton’s 2nd Law | | L = vt (2.1) q E a = | | ) m 1 Vertical distance travelled ☛ y = at2 2

Wednesday, January 30, 19 32 Review everything for next class BUT don’t forget

Wednesday, January 30, 19 33 Wednesday, January 30, 19 34 Wednesday, January 30, 19 35 HOMEWORK Review these slidesVector B4 Algebra watching superbowl 1.1 Definitions A vector consists of two components magnitude and direction (e.g. force, velocity,) pressure)

A scalar consists of magnitude only (e.g. mass, charge, density)

Euclidean vector, a geometric entity endowed with magnitude and direction as well as a positive-deﬁnite inner product; an element of a Euclidean vector space! In physics, Euclidean vectors are used to represent physical quantities that have both magnitude and direction, such as force, in contrast to scalar quantities, which have no direction

Wednesday, January 30, 19 36 Chapter 1

Vector Algebra

1.1 Deﬁnitions

A vector consists of two components: magnitude and direction . (e.g. force, velocity, pressure) A scalar consists of magnitude only. (e.g. mass, charge, density) 1.2 Vector Algebra 1.2 Vector Algebra

~aFigure+~b 1.1:= Vector~b + algebra~a

~a +(~b + ~c)=(~a +~b)+~c ~a +~b = ~b + ~a ~a +(~c + d~)=(~a + ~c)+d~

Wednesday, January 30, 19 37 1.3. COMPONENTS1.3. COMPONENTS OF VECTORSOF VECTORS 2 2

1.31.3 Components Components of of Vectors Vectors

UsuallyUsually1.3 vectors vectors Components are expressed are expressed according according toofcoordinate to coordinateVectors system system.. Each Each vector vector can can be expressedbe expressed in terms in terms of components of components. . Usually vectors are expressed according to coordinate system The mostTheEach commonmost vector common coordinatecan coordinatebe expressed system: system: Cartesianin Cartesianterms of components! The most common coordinate system Cartesian

~a = ax + ay + az ~a = ~a + ~a + ~a ~a = ~a x+ ~a y+ ~a z Magnitudex y ofz ~a = ~a = a Magnitude of ~a = ~a = a, Magnitude of ~a = ~a| |= a, | | | | 2 2 2 a =2 2ax +2 ay + az a = ax + ay + az 2 2 2 a = qax +qay + az q

~a = ax + ay 2 2 a = ax + ay

~a = ~ax + ~ay ax = qa cos 2 2 ~a a==~ax +ax~a+y ay ax =ayaq cos2=¡;a2asiny = a sin¡ a = ax + ay ay tan¡tan= q = a /a ax = a cosa ¡; ay = xa sin¡ a x Figure 1.2: ¡ measured anti-clockwise tan¡ = y Wednesday,from position Januaryx-axis 30, 19 ax 38 Figure 1.2: ¡ measured anti-clockwise from position x-axis Unit vectors have magnitude of 1 ~a aˆ = = unit vector along ~a direction Unit vectors have magnitude~a of 1 | | ~a aˆ = ˆi= unitˆj kˆ vectorare unit along vectors~a direction along ~a | | xyzllldirections ˆi ˆj kˆ are unit vectors along

lll ˆ ˆ ˆ xyz~adirections= ax i + ay j + az k

ˆ ˆ ˆ Other coordinate systems:~a = ax i + ay j + az k

Other coordinate systems: Unit vectors have magnitude of 1 ~a aˆ = = unit vector along ~ a direction ~a | |

ˆı |ˆ kˆ are unit vectors along

xyz directions

~a = ax ˆı + ay |ˆ + az kˆ

Wednesday, January 30, 19 39 1.3. COMPONENTS OF VECTORS 3 1. Polar Coordinates 1. Polar Coordinate:

~a = ar rˆ + a✓ ✓ˆ

~a = ar rˆ + aµ µˆ

Figure 1.3: Polar Coordinates

2. CylindricalWednesday, January 30, Coordinates: 19 40

~a = ar rˆ + aµ µˆ + az zˆ

rˆ originated from nearest point on z-axis (Point O’)

Figure 1.4: Cylindrical Coordinates

3. Spherical Coordinates:

~a = ar rˆ + aµ µˆ + a¡ ¡ˆ

rˆ originated from Origin O

Figure 1.5: Spherical Coordinates 1.3. COMPONENTS OF VECTORS 3

1. Polar Coordinate:

~a = ar rˆ + aµ µˆ

Figure 1.3: Polar Coordinates

2. Cylindrical2. Cylindrical Coordinates: Coordinates

! ~a = ar rˆ + aµ µˆ + az zˆ ~a = ar rˆ + a✓ ✓ˆ + az zˆ rˆ originated from nearest point on rˆoriginated from nearest pointz-axis on z-axis (Point (Point O’) O’)!

Figure 1.4: Cylindrical Coordinates

3. Spherical Coordinates:

Wednesday, January 30, 19 41

~a = ar rˆ + aµ µˆ + a¡ ¡ˆ

rˆ originated from Origin O

Figure 1.5: Spherical Coordinates 1.3. COMPONENTS OF VECTORS 3

1. Polar Coordinate:

~a = ar rˆ + aµ µˆ

Figure 1.3: Polar Coordinates

2. Cylindrical Coordinates:

~a = ar rˆ + aµ µˆ + az zˆ

rˆ originated from nearest point on z-axis (Point O’)

Figure 1.4: Cylindrical Coordinates 3. Spherical Coordinates 3. Spherical Coordinates:

! ~a = a rˆ + a µˆ + a ¡ˆ ~a = ar rˆ + a✓ ✓ˆr + aµ ˆ ¡ rˆoriginatedrˆ originatedfrom Origin fromO Origin O

Figure 1.5: Spherical Coordinates

Wednesday, January 30, 19 42 1.4 Multiplication of Vectors 1. Scalar multiplication If ~b = m~a ~b,~a are vectors; m is a scalar then b = ma (Relation between magnitude)

bx = max Components also follow relation by = may }

i.e. ~a = ax ˆı + ay |ˆ + az kˆ

m~a = max ˆı + may |ˆ + maz kˆ

Wednesday, January 30, 19 43 2. Dot Product (Scalar Product) Cont’d

ˆı ˆı = ˆı ˆı cos 0 =1 1 1=1 · | || | · ·

ˆı |ˆ = ˆı |ˆ cos 90 =1 1 0=0 · | || | · ·

ˆı ˆı =ˆ| |ˆ = kˆ kˆ =1 · · · ˆı |ˆ =ˆ| kˆ = kˆ ˆı =0 · · ·

If ~a = ax ˆı + ay |ˆ + az kˆ

~b = bx ˆı + by |ˆ + bz kˆ

then ~a ~b = a b + a b + a b · x x y y z z 2 ~a ~a = ~a ~a cos 0 = a a = a · | |·| | ·

Wednesday, January 30, 19 44 1.4. MULTIPLICATION OF VECTORS 5

3. Cross Product (Vector Product): 2. Cross Product (Vector Product)

If ~c = ~a ~b ⇥ If ~c = ~a ~b, £ then c = then~c =c =ab~c =sina b sin¡ | | | | ~a ~b = ~b ~a !!! ~a ~b £= 6 ~b £ ~a !!! ⇥ ~a6 ~b = ⇥~b ~a £ ° £ ~a ~b = ~b ~a ⇥ ⇥ Figure 1.7: Note: How angle ¡ is measured • Direction of cross product determined from right hand rule Direction of cross product determined from right hand rule. • Also,• ~ a ~ b is to ~a and ~b Also, ~a ~b is to ~a and ~b, i.e. • ⇥ £ ?? i.e. ~a (~a ~b)=0 ~a (~a ~b)=0 · £ · ⇥ ~b (~a ~b)=0 ~b (~a ~b)=0 · £ IMPORTANT: • · ⇥

~a ~a = a a sin0± =0 Wednesday, January 30, 19 £ · 45

î î = î î sin0± =1 1 0=0 | £ | | || | · · î ˆj = î ˆj sin90± =1 1 1=1 | £ | | || | · ·

î î = ˆj ˆj = kˆ kˆ =0 £ £ £ î ˆj = kˆ; ˆj kˆ = î; kˆ î = ˆj £ £ £

ˆi ˆj kˆ ~ ˆ ~a b = ax ay az = (ay bz az by) i £ Ø Ø ° Ø b b b Ø +(a b a b ) ˆj Ø x y z Ø z x ° x z Ø Ø +(a b a b ) kˆ Ø Ø x y ° y x 1.4. MULTIPLICATION OF VECTORS 5

3. Cross Product (Vector Product):

If ~c = ~a ~b, £ then c = ~c = a b sin¡ | | ~a ~b = ~b ~a !!! £ 6 £ ~a ~b = ~b ~a £ ° £

Figure 1.7: Note: How angle ¡ is measured

Direction of cross product determined from right hand rule. • Also, ~a ~b is to ~a and ~b, i.e. • £ ? ~a (~a ~b)=0 · £ ~b (~a ~b)=0 · £ IMPORTANT: •

~a ~a = a a sin0± =0 £ ·

• IMPORTANT ~a ~a = a a sin 0 =0 ⇥ · ˆ ˆ ˆ ˆ i i = ˆıi iˆı =sinˆı 0ˆı±sin=1 0 =11 0=01 0=0 | £ | || ⇥|| || | || | · · · ˆi ˆj = ˆi ˆj sin90± =1 1 1=1 | £ | ˆı| |||ˆ |= ˆı |ˆ sin 90 =1· · 1 1=1 | ⇥ | | || | · ·

ˆı |ˆ kˆ ~a ~b = a a a =(a b a b )ˆı +(a b a b )ˆ| +(a b a b ) kˆ ⇥ x y z y z z y z x x z x y y x ˆ ˆ b ˆb b i j x k y z ~ ˆ ~a b = ax ay az = (ay bz az by) i £ Ø Ø ° Ø bWednesday,b Januaryb 30, 19 Ø +(a b a b ) ˆj 46 Ø x y z Ø z x ° x z Ø Ø +(a b a b ) kˆ Ø Ø x y ° y x 4. Vector identities

~a (~b + ~c )=~a ~b + ~a ~c ⇥ ⇥ ⇥ ~a (~b ~c )=~b (~c ~a )=~c (~a ~b) · ⇥ · ⇥ · ⇥

~a (~b ~c )=(~a ~c )~b (~a ~b)~c ⇥ ⇥ · · 1.5 Vector Field (Physics Point of View)

A vector field ~ ( x, y, z ) is a mathematical function F which has a vector output for a position input

(Scalar ﬁeld ( x, y, z ) ) U

Wednesday, January 30, 19 47 1.5. VECTOR FIELD (PHYSICS POINT OF VIEW) 6

4. Vector identities:

~a (~b + ~c)=~a ~b + ~a ~c £ £ £ ~a (~b ~c)=~b (~c ~a)=~c (~a ~b) · £ · £ · £ ~a (~b ~c)=(~a ~c)~b (~a ~b)~c £ £ · ° · 1.5 Vector Field (Physics Point of View)

A vector ﬁeld ~ (x, y, z) is a mathematical function which has a vector output F for a1.5.position VECTORinput. FIELD (PHYSICS POINT OF VIEW) 6

(Scalar4. Vector ﬁeld identities:~(x, y, z)) U ~a (~b + ~c)=~a ~b + ~a ~c £ £ £ ~a (~b ~c)=~b (~c ~a)=~c (~a ~b) · £ · £ · £ 1.6 Other Topics~a (~b ~c)=(~a ~c)~b (~a ~b)~c £ £ · ° · Tangential Vector 1.5 Vector Field (Physics Point of View)

A vector ﬁeld ~ (x, y, z) is a mathematical function which has a vector output for a position input.F

(Scalar ﬁeld ~(x, y, z)) U 1.6 Other Topics 1.6 Analytic Geometry Tangential Vector Tangential Vector Figure 1.8: d~l is a vector that is always tangential to the curve C with inﬁnitesimal length dl

Surface Vector Surface Vector Figure 1.8: d~l is a vector that is always tangential to the curve C with inﬁnitesimal length dl

Surface Vector

Figure 1.9:Somed~a uncertainty!is a vector that ( d ~a is alwaysversus perpendicular d ~a ) to the surface S with inﬁnitesimal area da

Wednesday, January 30, 19 48 Figure 1.9: d~a is a vector that is always perpendicular to the surface S with inﬁnitesimal area da 1.6. OTHER TOPICS 7

1.6. OTHER TOPICS 7 Some uncertainty! (d~aversus d~a) °

Two conventions: Some uncertainty! (d~aversus d~a) ° Area formed from a closed curve • TwoTwo conventions: conventions: Area formed from a closed curve • • Area formed from a closed curve

Figure 1.10: Direction of d~a determined from right-hand rule

Closed surface enclosing a volume • • ClosedFigure 1.10: surface Direction enclosing of d~a determined a volume from right-hand rule

Closed surface enclosing a volume •

Wednesday,Figure January 1.11:30, 19 Direction of d~a going from inside to outside 49

Figure 1.11: Direction of d~a going from inside to outside