Physics II Reviewer.pdf

7/22/2019 Physics II Reviewer.pdf

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STATIC ELECTRICITY

Coulomb's Law

F =(K |q1q2|)

r2

where: q1& q 2= electric charges in coulomb, C

r = distance between two charges

K = 8.99 x 109N m

2/ C

2

F = electric force in newton, N

in CGS units:

K= 1 dyne cm2

/ statC2

r = cm

F = dyne

other units:

1 C = 3 x 109

statC

1 N = 1 x 105dynes

1 C = 10-6

C

1 nC = 10-9

C

1 C = 10-12

C

Determination of Direction of F

Like charges, REPEL

Unlike charges, ATTRACT

Resultant Force

vector sum of electric forces in a system ofcharges

GAUSS' LAW

Electric Flux (lines of force) produced at a spot normal

to the surface where electric field exists due to electric

charge q.

= q

for spherical surface with area, An = 4R2

= oE An

where:

o = 1 / 4K = 8.85 x 10-12

C2

/ N m2

UNIFORM ELECTRIC FIELD

E =F

q =

ma

q

y = Vot + at2

H = (Vosin)2

/ 2a

tH= V osin/ a

R = (Vocos) (tR)

SAMPLE PROBLEMS:

1. Two particles with electric charges Q and 3Q areseparated by a distance of 1.2 m. (a) If Q 4.5 C, what is

the electric force between the two particles? (b) If Q 4.5

C, how does the answer change?

2. Two small, identical particles have charges Q1 3.0 mCand Q2 5.0 mC. If the electric force between the particles

is 120 N, what is the distance between the particles?

Answer: 0.034 m3. Three charges with q 7.5 mC are located in the positive

and negative x-axis and in the negative y-axis, with L=25

cm from the origin. (a) What are the magnitude and

direction of the total electric force on the charge at the

bottom? (b) What are the magnitude and direction of the

total electric force on the charge at the right?

Answer: 5.7 N downward (b) 5.7 N at 30.4 above horizontal

4. Particles of charge Q and 3Q are placed on the x axis ax= -L andx= + L, respectively. A third particle of charge q

is placed on the x axis, and it is found that the tota

electric force on this particle is zero. Where is the

particle?

Answer: Atx = 0.27L

5. What is the magnitude of the electric field at a distanceof 1.5 m from a point charge with Q = 3.5 C?

Answer: 1.4 x 1010

N/C

6. A charged paint is spread in a very thin uniform layerover the surface of a plastic sphere of diameter 12 cm

giving it a charge of -15C. Find the electric field (a) just

inside the paint layer; (b) just outside the paint layer; (c

5cm outside the surface of the paint layer.

Answer: (a) 0 (b) 3.75 x 107N/C radially inward (c) 1.11 x

107N/C radially inward

ELECTRIC POTENTIAL

Due to an isolated charge qV =

Due to two or more point charges q1, q2, q3V = V1+ V2+ V3+

V = k Potential energy due to two charges Q and q

PE = Fr = (J, ergs)

Note: Include signs of charges in computation

PE due to three chargesPE = kq

Example:


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1.) Two point charges +40nC and -30nC are 10 cm apart. Point

A is midway between the charges and point B is 6 cm from

+40nC and 8 cm from +30 nC.

a. Electric potential at point A

b. Electric potential at point B

SOLUTION:

a. Va(+40)=9109(40 109)

0.05

= 7200 volts

Va(-30)=(9 109) (30 109)

0.05= -5400 volts

Va= 7200 5400

Va= 1, 800 volts

b. Vb(+40)=9 109(40 109)

0.06= 6000 volts

Vb(-30)=9 109(30 109)

0.08= -3375 volts

Vb= 6000 3375

Vb= 2625 volts

POTENTIAL DIFFERENCE

= Va- VbVab= Va- VbVba= Vb- Va

Vab= -Vba

* Referring to example 1, determine the potential difference

between A and B, Vab

Vab= V a- V b= 18002625

Vab= -825 volts

WORK DONE, W

W = q Signs of W:

W is + :Va>Vb

W is -: Va


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a. Va(12) =9 109(12 109)

0.06= 1800 V

Va(-12)=9 109(12 109)

0.04= -2700 V

Va= -900 V

Vb(12) = 9 109(12 109)0.04

= 2700 V

Vb(-12)=9 109(12 109)

0.14= -771. 43 V

Vb= 1928.57 V

Vc(12) =9 109(12 109)

0.1= 1080 V

Vc(-12) =9 109(12 109)

0.1= -1080 V

Vc = 0V

b. EPEa= (4 x 10-9

) (-900) = -3.6 x 10-6

J

EPEb= (4 x 10-9

) (1928.57) = 7.71 x 10-6

J

EPEc = (4 x 10-9

) (0) = 0

c. Vab= (-900 V) (1928.57) = -2828.57 V

Vba= 2828.57 V

Vbc= (1928.57) (0) = 1928.57 V

Vcb= -1928.57 V

d. Wab= (4 x 10-9

) (-2828.57) = 1.13 x 10-5

J

Wcb= (4 x 10-9

) (-1928.57) = -7.71 x 10-6

J

DC CIRCUITS

Electric Current I = q/t Ampere (A)

Resistance R = V/I Ohms

Voltage V = IR Volts (V)

Power P = IV

P = I2R

P = V2

/R

Watts (W)

Energy W = Pt = JQ Joule (J)

Heat Q = Pt/J

Q = 0.24 VIt

Cal

MEH 4.186 J/cal J/cal

Examples:

1. What is the current when a charge of 15 coulombs flows in

a circuit in 1.5 seconds?

I = q/t

I = 15/1.5

I = 10 A

2. How much heat is developed in one minute in an electric

heater which draws a current of 5 A when connected to a 110V line?

Q = 0.24 VIt

Q = (0.24) (110) (5) (60 s)

Q = 7920 calories

SERIES CIRCUIT

I is the same V is subdivided R is the sum of Ris

PARALLEL CIRCUIT

I is subdivided V is the same R is less than the lesser/least value of Rs

Example

1.) In the circuit, let R1= 5 , R2= 10 , R3= 4 and V = 2V.Find the currents I1, I2, I3.

V12= V 1= V 2R = R12 + R 3

R12=5(10)

15= 3.33

R = 3.33+ 4 = 7.33

I = I12= I 3

I = 2/3.33 = 0.27 A

I3= 0.27 A

V12= I 12R12

= (0.27 A) (3.33)

= 0.9 V

V1= 0.9 V

2 V

5

10

4


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i

`1R1 `2R2

`1R1

`2R2

A C

DB

A

B

C

D

V12= 0.9 V

I1= 0.9 / 5 = 0.18 A

I2= 0.9 / 10 = 0.09 A

2.) In the circuit, (a) Find the currents through each resistor.

R1= 3 , R2= 6, R3= 6 , R4= 12 , V = 18 V. (b) What is thetotal current i?

I = I12= I 34

V12= V 1=

V2V34= V 3=

V4R = R12 +

R34

R12 = (3 x 6) / 9 = 2 R34= (6 x 12) /18 = 4

R = 6

I = V/R = 18 V/6I = 3 A

V12= I 12R12= 3 (2) = 6V

V34= I 34R34= 3(4) = 12 V

I1= 6V / 3 = 2 AI2= 6 V / 6 = 1 AI3= 12V / 6 = 2 AI4= 12V / 12 = 1 A

BATTERY WITH EMF

Vab= terminal voltage

r = internal resistance

Example:

Determine:

a) Current delivered by the battery

b) Terminal voltage of the battery

c) Power delivered by the battery

a. I =

+ =12

1+5= 2 A, clockwise

b. Vab= Ir= 122r

= 10 V

c. P = IV

= 12 (2)

= 24 watts

2 OR MORE EMFS:

Series circuit (aiding)

I =+

+++

Series Circuit (opposing)

I =

+++

V

R3R1

R2 R4

R = 512 V

1


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loop

loop

loop

loop

Example:

R1= 4 , R2= 6 1= 12 V, 1 2= 6 V, 0.5

Find:

a. Current in the circuit and its direction

b. Terminal voltage

*Refer to the figures above

*Series circuit opposing

I =126

1+ 0.5+4+6= 0.52 A, clockwise

Vab= 12 (0.52) = 11.48 V

Vcd= 6 + 0.52 (0.5) = 6.26 V

*Series circuit aiding

I =12+6

1+0.5+4+6= 1.57 A, clockwise

Recall: Terminal Voltage

(potential difference between 2 points)

Vab= E 1Ir V ab= E 2+ Ir

Where E = emf

Ir = voltage drop

Example:

The current I through R1in the circuit diagram is 40 mA. (a)

What is the current through R2, R3and R 4? (b) What is thepotential difference between A and B?

0.04 =21.5

2+3.75+4

V2= 1.89 V

V23= I 23R23

V23= (0.04) (3.75)

V23= 0.15 V

V23= V 2= V 3I2= V 2/ R 2= 0.15 / 5 = 0.03 A

I3= 0.15 / 15 = 0.01 A

Vba= 1.5 + 0.04 (3.75)= 1.65 V

Vab= -1.65 V

KIRCHHOFFS RULES

KVL:

V = 0

KCL:

I = 0

LOOP DIRECTION:

+E

-E

-IR

+IR

Example:


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The voltage drop across R3in the circuit diagram is 4 V. (a)

Find the currents through the resistor R1, R2and R 3. (b) What

is the resistance of R2?

V = 0

Loop ACDA

6i 1R1i 3R3= 0

64i (0.8)(0.5) = 0

I = (4-6) / 4

I1= 0.5 A

Loop ABCA

1.5 i 2R2i 1R1= 0

1.5 i 2R2= 2

I2R2= 0.5

I = i1+ i 2

I2= i i 1I2= 0.8 0.5 = 0.3 A

R2= 1.67 SAMPLE PROBLEMS:

1.) A charge of +16nC is fixed at the origin of coordinates. A

second charge of +12nC is at x= 4m, y = 0 and a third

unknown charge is at x = 11 m,y = 0. If the resultant electric

field at x = 8 m, y = 0 is 24.25 N/C directed to the right:

a. What is the magnitude of the electric field at x = 8m, y = 0

due to the +16nC charge?

b. What is the magnitude of the elctric field at x = 8m, y=0

due to the +12nC charge?

c. What are the sign and magnitude of the third charge?

d. How far from the origin should the third charge be placed

so that the resultant electric field at x= 8m, y=0 is zero?

2.) A charge of +20nC is 20 cm from a second charge of -20nC.

Point A is 12 cm from the 20nC charge and 8 cm from the -

20nC. Point B is 8cm from the 20 C charge and 28 cmfrom the

-20C. Point C is 20 cm from the 20nC chrge and 20 cm from

the -20nC.

a. What is the potential difference Vab?

b. What is the potential difference Vbc?

c. How much work is done on the electric field I moving a

charge of 4nC from A to B?

d. How much work is done by the electric field in moving a

charge of 4nC from B to C?

3. I the circuit shown, R1= 1 , R2= 10 ,I 1= 3.5 A and I 4= 2A.The battery has negligible internal resistance. If the rate at

which R4is dissipating electrical energy is 20 W;

a. What is R4?

b. What is R3?

c. What is I3?

d. What is the emf of the battery?

4. A uniform electric filed of 5000 N/C exists in the region

between two oppositely charged parallel plates separated by

a distance of 0.05 m. An electron is released from rest at the

surface of the negative plate.

a. What is the magnitude of the force experienced by the

electron as it moves toward the positive plate?

b. What is the work done against the electric field in moving

the electron across the plates?

c. What is the poential difference between the plates?d. What is the speed of the electro as it strikes the opposite

plate?

5. If the internal resistances of the batteries in the circuit

below are

r1= 0.1 ,r2= andr3= 0.3 respective

ly.

a. What is

the

current

through

the 3 V

battery?

b. What is the current through the 8 V battery?

c. What is the current through the 12 V battery?

d. What is the terminal voltage of the 3 V battery?

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Physics II Reviewer.pdf