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PHYS-2020: General Physics II

Course Lecture Notes

Section I

Dr. Donald G. Luttermoser East Tennessee State University

Edition 3.

Abstract

These class notes are designed for use of the instructor and students of the course PHYS-2020: General Physics II taught by Dr. Donald Luttermoser at East Tennessee State University. These notes make reference to the College Physics, Enhanced 7th Edition (2006) textbook by Serway, Faughn, and Vuille.

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B. Insulators and Conductors.

1. Conductors are materials in which electric charges move freely (i.e., they have low internal resistance ). a) Copper.

b) Aluminum.

c) Silver.

2. Insulators are materials in which electric charges do not freely move (i.e., they have high internal resistance ). a) Glass.

b) Rubber.

3. Semiconductors are materials that lie in between these other two =⇒ if controlled amounts of foreign atoms are added to semi- conductors, their electrical properties can be changed by orders of magnitude. a) Silicon.

b) Germanium.

4. The Earth can be considered to be an infinite reservoir of (or for) electrons. a) It can accept or supply an unlimited number of electrons.

b) When a conductor is connected to the Earth (e.g., con- ducting wire or copper pipe), it is said to be grounded =⇒ lightning rods.

Donald G. Luttermoser, ETSU I–

5. An object can be charged in one of two ways: a) Conduction: Charge exchange through contact. i) If one charged object comes in contact with as sec- ond object, charge can move from the charged ob- ject to the uncharged object.

ii) Rubbing two different materials together, called frictional work, can produce negative charge on one object and a positive charge on the other.

b) Induction: Charge exchange with no contact. i) Charge one object.

ii) This charge produces an electric field to form.

iii) This electric field can then induce charges to mi- grate on a second object =⇒ the second object po- larizes (see Figure 15.4 in the textbook).

C. Coulomb’s Law.

1. Coulomb’s law states that two electric charges experience a force between them such that:

a) It is inversely proportional to the square of the separation r between the 2 particles along the line that joins them.

b) It is proportional to the product of the magnitudes of the charges, q 1 and q 2 , on the two particles.

c) It is attractive if the charges are of opposite sign and re- pulsive if the charges have the same sign.

Donald G. Luttermoser, ETSU I–

Example I–1. Problem 15.3 (Page 525) from the Serway & Faughn textbook: An alpha particle (charge +2. 0 e) is sent at high speed toward a gold nucleus (charge +79e). What is the elec- trical force acting on the alpha particle when it is 2. 0 × 10 −^14 m from the gold nucleus? Solution: Use Coulomb’s Law (i.e., Eq. I-1) where q 1 = +2e, q 2 = +79e, r = 2. 0 × 10 −^14 m, and e = 1. 60 × 10 −^19 C:

| F~e| = ke^ |q^1 | |q^2 | r^2 = ke^ (2e)(79e) r^2 =

  8. 99 × 109 N^ ·^ m

2 C^2

 

  (2.^0 ·^ 79)(1.^60 ×^10

− 19 C) 2

(2. 0 × 10 −^14 m)^2

 

= 91 N (repulsion).

The Coulomb force is a repulsion force here since both particles have charges of the same sign.

D. The Electric Field.

1. An electric charge emits an electric field which always points away from a positive charge and points towards a negative charge:

E-field points away

−

E-field points to charge

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a) Assume we take a small (i.e., negligible) positive test charge q◦, and place it near a larger positively charged object, Q.

Q

qo

E

b) The direction of E~ (i.e., the electric field) at a point is defined to be the direction of the electric force that would be exerted on a small (i.e., negligible with respect to the charge on the larger object, q◦  Q) positive charge placed at that point.

c) Note that if q◦ starts to become comparable in size (in terms of charge) to Q, the electric field of Q will be altered by q◦.

2. Hence, the electric field is defined by the electric force exerted on a charged particle by another charged particle or object:

E^ ~ =

F~e q◦

. (I-3)

a) Note that from this equation, E always points in the same direction that the force exerted on a positively charged particle q◦ by a charged object that gives rise to E~.

b) E is measured in N/C in SI units and dyne/esu in the cgs system.

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electric field, it experiences a retarding force given by Eq. (I-3): F = −eE , negative since the E-field slows the electron. Since this force slows the electron, it produces a deceleration. Using Newton’s 2nd law we can write

a = F me = −eE me

= −(1.^60 ×^10

− 19 C)(1000 N/C)

9. 11 × 10 −^31 kg = − 1. 76 × 1014 m/s^2.

Using one of the 1-D equations of motion from General Physics I, we can solve for the distance that the electron travels before coming to a stop, ∆x. We have v = 0 (e−^ comes to rest), v◦ =

3. 00 × 106 m/s, and a has been calculated above, giving v^2 = v^2 ◦ + 2a ∆x ∆x = v

(^2) − v (^2) ◦ 2 a = 0 −^ (3.^00 ×^10

(^6) m/s) 2 2 .0(− 1. 76 × 1014 m/s^2 ) = 0 .0256 m = 2 .56 cm.

4. When summing E-fields from multiple charges, it is best to draw a vector diagram first at a point where the E-field is to be deter- mined. Draw the E-field vector of each charge in the direction as dictated by the rules above. Then vectorially add the individual E-fields together, letting the diagram determine the sign of the E-field (typically + to the right and + upwards). Then add using the absolute value of the charge:

E = ∑N 1

(±)ke|q| r^2

, (I-5)

Donald G. Luttermoser, ETSU I–

where either + or − is selected based upon the direction of the E-filed vector (see Example I-3 below).

Example I–3. Positive charges are situated at three corners of a rectangle as shown in the figure below. Find the electric field at the fourth corner.

q 2 = 2.00 nC q 3 = 5.00 nC

q 1 = 6.00 nC E (^1)

E 2

E 3

r 1 = 0.600 m

r 2 = 0.200 m

r 3

φ

φ

Solution: From the geometry of the rectangle as shown in the figure above, we have

r 32 = r 12 + r^22 = (0.600 m)^2 + (0.200 m)^2 = 0.400 m^2.

and φ = tan−^1

(r 2 r 1

) = tan−^1

( (^0) .200 m 0 .600 m

) = 18. 4 ◦^.

The components of the individual E-field vectors are

E^ ~ 1 = −E 1 x ˆx = −E 1 xˆ E~ 2 = +E 2 y yˆ = +E 2 yˆ E~ 3 = −E 3 x ˆx + E 3 y yˆ = −E 3 cos φ xˆ + E 3 sin φ y .ˆ

The resultant E-field at the vacant corner is

E^ ~R = E~ 1 + E~ 2 + E~ 3

Donald G. Luttermoser, ETSU I–

E. Electric Field Lines.

1. Drawing electric field lines: a) The electric field vector, E~, is tangent to the electric field lines at each point.

b) The number of lines per unit area through a surface ⊥ to the lines is proportional to the strength of the electric field in a given region. i) E~ is large when the field lines are close together.

ii) E~ is small when the field lines are far apart.

2. For a system of charged particles, the following rules apply: a) The lines must begin on positive charges (or at infinity) and must terminate on negative charges (or, in the case of excess charge, at infinity).

b) The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the mag- nitude of the charge (e.g., if a charge of +q has n-lines per unit volume leaving the charge, a charge of − 2 q will have 2 n-lines per unit volume approaching the charge — see Ex. I-4).

c) No 2 field lines can cross each other.

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• −

2 opposite but equal charges

field stronger

weakerfield

• 2 equal charges +

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d) On an irregular shaped conductor, charge accumulates at its sharpest points.

charge accumulates here

conductor

G. The Electric Flux and Gauss’s Law.

1. The electric flux is a measure of the number of E-field lines that crosses a given area. a) An E-field whose lines penetrate a cross-sectional (or sur- face) area A ⊥ to A has an electric flux ΦE given by

ΦE = E A. (I-6)

b) However, if the E-field lines lie at an angle θ with re- spect to the normal line of area A (see Figure 15.25 in the textbook), the electric flux is given by the more general formula: ΦE = E~ · A~ = E A cos θ , (I-7)

where the ‘·’ is called the dot product operation and A^ ~ has magnitude A (the total cross-sectional or surface area) and direction given by the normal line of the area. The angle θ is the angle between that normal line and the E-field direction.

c) When the area is constructed such that a closed surface is formed, we shall adopt the convention that the flux lines

Donald G. Luttermoser, ETSU I–

passing into the interior of the volume are negative and those passing out of the interior of the volume are positive.

2. One of the most important law’s in all of electromagnetism is Gauss’s law: a) If we use a sphere for our enclosing volume, the sphere has a surface area of A = 4πr^2. If we place charge q at the center of this sphere, then the field lines will always be ⊥ to the surface of the sphere since they point radially outward. Using Eq. (I-4) in Eq. (I-6), we can write the electric flux as

ΦE = E A = ke q r^2

( 4 πr^2

) = 4πkeq ,

independent of the distance from the charge!

b) In electromagnetism, there is a constant related to the Coulomb constant called the permittivity of free space ◦. This constant is given by

◦ =

4 πke

= 8. 85 × 10 −^12

C^2

N · m^2

. (I-8)

c) Using this constant in our flux equation above, we can write the electric flux as

ΦE = 4πkeq = q ◦

d) Using calculus, we could show that this simple result is true for any closed surface (even non-symmetrical ones) that surrounds any charge q =⇒ such a surface is referred to as a gaussian surface.

e) This bit of mathematics was first worked out by Gauss and is therefore called Gauss’s law.

Donald G. Luttermoser, ETSU I–

angle to the x axis (and hence E field), so θ = 90◦^ and

ΦE = EA cos θ = (3. 50 × 103 N/C)(0.245 m^2 ) cos 90◦^ = 0.

Solution (c): Since the E field is along the x axis and the normal of the area make an angle of 40. 0 ◦^ with respect to the x axis, θ = 40. 0 ◦, so

ΦE = EA cos θ = (3. 50 × 103 N/C)(0.245 m^2 ) cos 40. 0 ◦ = 657 N · m^2 /C.