Physics 72.1 Practical Exam Reviewer

PHYSICS 72.1 REVIEW FOR PRACTICAL EXAM

2

ND

_{SEMESTER, A.Y. 2015-2016}

R. AGUILAR, N. CABELLO, D. LUMANTAS

NATIONAL INSTITUTE OF PHYSICS UNIVERSITY OF THE PHILIPPINES DILIMAN, QUEZON CITY 1101

ELECTRIC FIELD & EQUIPOTENTIAL LINES

Reference: Physics 72.1 Electric Field & Equipotential Lines Lab Manual 2015

ELECTRIC POTENTIAL DIFFERENCE



Δ𝑉: potential difference

Equipotential Lines (green)



Points in space that have the same

electric potential with respect to

the same reference point

Electric Field Lines (red)



Always perpendicular to an

equipotential line (see Eq. 4)



Electric field points towards

decreasing potential

Equipotential and electric field lines of 2 equal but oppositely charged particles [1]

[1] http://www.alpcentauri.info/equipotential_lines.html

1) WHICH POINT HAS THE LARGEST MAGNITUDE OF ELETRIC FIELD?

2) WHAT IS THE DIRECTION OF ELECTRIC FIELD AT THAT POINT?

A

B

C D

ANSWERS:

1) A. Equipotential lines are closest at that point.

2)  (to the right). Since electric field lines are always perpendicular to equipotential lines and they point towards decreasing potential.

OHM’S LAW

V: Voltage (V)

R: Resistance (Ω)

I: Current (A)

OHM’S LAW

PLOTS (y vs. x)

1)

Voltage vs. Current



Slope = R

2) Current (y) vs. 1/Resistance



Slope = V

3) Voltage vs. Resistance

 Slope = I

4) Resistance vs. Length

OHM’S LAW (LINEAR REGRESSION)

Example 1 (Voltage vs. Current)

y = 3.9x - 0.084 R² = 0.9994 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.1 0.2 0.3 0.4 0.5 0.6 Vo lt ag e (V) Current (A)

Voltage vs. Current

Current (A) Voltage (V)

0.1 0.3

0.2 0.7

0.3 1.09

0.4 1.4

0.5 1.82

3) USING THE GIVEN VALUES AND THE PLOT, WHAT IS THE VALUE OF

RESISTANCE OF THE MATERIAL?

Answer: Based from the slope of the

graph, 𝑹 = 𝟑. 𝟗 𝛀.

OHM’S LAW (LINEAR REGRESSION)

Example 2 (Resistance vs. Length of Wire)

y = 0.0094x + 1.1552 R² = 0.9973 0 0.5 1 1.5 2 2.5 3 3.5 0 50 100 150 200 250 R esist an ce (Ω ) Length (cm)

Resistance vs. Length of Wire

Length (cm) Resistance (Ω) 100 2.1 120 2.3 140 2.48 160 2.65 180 2.83 200 3.07

4) USING THE GIVEN VALUES AND THE PLOT, IF THE CROSS-SECTIONAL AREA IS 2 cm2_,

WHAT IS THE RESISTIVITY OF THE MATERIAL?

Answer:

𝜌 = 𝑠𝑙𝑜𝑝𝑒 ∗ 𝐴 = 0.0094 * 2 𝝆 = 𝟎. 𝟎𝟏𝟖𝟖 𝛀𝐜𝐦

CIRCUIT ANALYSIS



Circuit – conducting path where current can flow and the components that make up

this path.



Steady current – only possible for closed loops or complete circuits with at least one source of

electromotive force (emf) that supplies electrical energy to the circuit.



For circuits composed of resistors connected in both series and/or parallel, Ohm’s law applies:

𝑽 = 𝑰𝑹

_𝒆𝒇𝒇

CIRCUITS IN SERIES AND PARALLEL



In Series:

𝑅𝑒𝑓𝑓 = 𝑅

₁

+ 𝑅

₂

+ 𝑅

₃

+ …+𝑅

_𝑁



In Parallel:

1

𝑅

_𝑒𝑓𝑓

=

1

𝑅

₁

+

1

𝑅

₂

+

1

𝑅

₃

+ ⋯ +

1

𝑅

_𝑁

EXAMPLE CIRCUIT – FROM CLASS

𝑅𝑒𝑓𝑓 = 𝑅

₁

+ 𝑅

₂

||𝑅

₃

= 𝑅

₁

+

1

𝑅

₂

+

1

𝑅

₃ −1

= 𝑅1 +

𝑅

2

𝑅

3

𝑅

₂

+𝑅

₃

R

₂

and R

₃

are in parallel (R

₂

||R

₃

)

KIRCHOFF’S RULES



Loop Rule



σ

_𝑖

𝑉

_𝑖

= 0



Sum of changes along a closed path is zero



Junction Rule



σ 𝐼

_{𝑒𝑛𝑡𝑒𝑟𝑖𝑛𝑔}

= σ 𝐼

_{𝑙𝑒𝑎𝑣𝑖𝑛𝑔}



Current in = current out

http://www.wikipremed.com/01physicscards.php?card=708

5) SET-UP JUNCTION RULE AT PT. P AND LOOP RULE FOR A AND B.

A

B

P

I

₃

I

₁

I

₂

Answers: Loop A: 𝑉₁ − 𝐼₁𝑅₁ − 𝐼₂𝑅₂ = 0 Loop B: 𝑉₂ − 𝐼₂𝑅₂ = 0 Junction P: 𝐼₁ + 𝐼₃ = 𝐼₂:

Outer Loop (redundant with Loops A and B):

MEASURING PARAMETERS USING MULTIMETER

Measuring voltage across resistor Measuring current through circuit Measuring resistance of resistor

CAPACITOR



Device that stores electrical energy

https://en.wikipedia.org/wiki/Capacitor

𝐶 =

𝑞

𝑉

Capacitance:

𝑞 −

𝑉 −

Charge stored in the capacitor

Potential difference between the

capacitor

RC CIRCUIT

Discharging:

Charging:

𝜏 = 𝑅𝐶

6) WHAT IS THE TIME CONSTANT OF THE RC CIRCUIT BELOW?

7) WHAT IS ITS VOLTAGE AT t = 4RC?

100 𝑀Ω

1000 𝑛𝐹

Answers: Time constant 𝜏 = 100𝑥106 Ω ∗ 1000𝑥10−9 F 𝝉 = 𝟏𝟎𝟎 𝒔 Voltage at t=4RC 𝑉 𝑡 = 𝑉_𝑜 1 − 𝑒−𝑡/𝑅𝐶 𝑉 = (5.5 𝑉) 1 − 𝑒−4𝑅𝐶/𝑅𝐶 𝑉 = 5.5𝑉 1 − 𝑒−1 𝑉 = 0.63 5.5 𝑉 𝑽 = 𝟓. 𝟑𝟗𝟗 𝑽 = 𝟓. 𝟒𝟎 𝑽 5.5V CHARGING

MAGNETIC FLUX

A

B

B











http://ibphysicsstuff.wikidot.com/electromagnetic-induction

Magnetic flux changes by

changing:



the magnitude of the

magnetic field



changing the surface area



changing the relative

orientation of the field and

the surface normal

FARADAY’S LAW OF INDUCTION



The induced emf in a closed loop equals the negative of the

time rate of change of the magnetic flux through the loop

dt

d



_B







Lenz’s Law:

An induced current will be in such a direction as to produce

a magnetic field that will

oppose

the motion of the magnetic

field that is producing it.

INDUCTION EXPERIMENT 1

http://voer.edu.vn/c/faradays-law-of-induction-lenzs-law/0e60bfc6/99a3eaad

Magnet

Actual Direction

I

induced

B

induced

North pole moves toward solenoid cw right South pole moves towards soleniod ccw left North pole moves away from solenoid ccw left South pole moves away from solenoid cw right

INDUCTION EXPERIMENT II



Magnetic Field of a solenoid

𝐵 = 𝜇𝑛𝐼



Magnetic permeability of air and aluminum: almost equal to 𝜇

₀



Magnetic permeability of iron > 𝜇

₀

Magnetic

INDUCTION EXPERIMENT III

overlap

Increasing deflection upon

turning on and turning off

8) WHAT IS THE MAGNETIC PERMEABILITY FOR N = 50 TURNS OF A

1.0-METER SOLENOID WITH THE FOLLOWING PLOT OF B VS. I?

y = 3.9x - 0.084 R² = 0.9994 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.1 0.2 0.3 0.4 0.5 0.6 M ag n etic F iel d ( T) Current (A)

Magnetic Field vs. Current

Current (A) Magnetic Field (B)

0.1 0.3 0.2 0.7 0.3 1.09 0.4 1.4 0.5 1.82 Answer: 𝑛 = 𝑁 𝐿 = 50 𝑡𝑢𝑟𝑛𝑠 1.0 𝑚 𝝁 = 𝒔𝒍𝒐𝒑𝒆/𝒏 = 𝟎. 𝟎𝟕𝟖 𝑻𝒎/𝑨.

SOURCES OF MAGNETIC FIELD

 A wire carrying current produces a magnetic field (direction determined by right-hand rule)

 When the wire is looped, the field near the center becomes perpendicular to the direction of the loop  Multiple loops increase the field strength – solenoid

𝐵 = 𝜇0𝐼 2𝜋𝑟 𝐵 = 𝜇0𝑁𝐼 𝐿 𝐵 = 𝜇₀𝑛𝐼 𝜇₀ = 4𝜋 × 10−7𝑇 ∙ 𝑚/𝐴

Magnetic permeability of vacuum Wire is

looped

Multiple loops

SOURCES OF MAGNETIC FIELD

 Addition of material inside solenoid modifies magnetic permeability (µ₀ becomes

µ=kµ₀)

 Increase in magnetic permeability results to increase in B strength

MAGNETIC FIELD LINES FROM DIFFERENT

CONFIGURATIONS

Single bar magnet

Two bar magnets

unlike poles facing each other

Two bar magnets like poles facing each other

SUPERPOSITION OF WAVES

 Consider two waves travelling through the same medium at the same time.

The net displacement of the medium at any point in space or time, is simply the sum of

the individual wave

displacements

Interference: combination of

two or more waves to form a composite wave

INTERFERENCE

DIFFRACTION

SINGLE SLIT EXPERIMENT

𝑎 sin 𝜃 = 𝑚𝜆 (1) sin 𝜃 = 𝑦𝑚,𝑑𝑖𝑓𝑓 𝐿 (2) a𝑦𝑚,𝑑𝑖𝑓𝑓 𝐿 = 𝑚𝜆 (3) 𝑎 – slit width 𝑦_{𝑚,𝑑𝑖𝑓𝑓} − mth intensity minimum 𝐿 − slit to screen distance

DOUBLE SLIT INTERFERENCE

Diffraction envelope y 𝑦_{𝑚,𝑖𝑛𝑡} = 𝑚𝜆𝐿 𝑑

Condition for maximum:

𝑎 – slit separation

𝑦_{𝑚,𝑖𝑛𝑡} − mth intensity peak from the center 𝐿 − slit to screen distance

9) GIVEN THE FOLLOWING FIGURE, WHAT IS THE WAVELENGTH OF

THE LIGHT SOURCE?

L = 1.0 m Δy_𝑚 = 4 mm d a a a = 0.04 mm d = 0.25 mm LIGHT SOURCE

Better to use single-slit diffraction

equation since the given Δ𝑦_𝑚 is at the dark fringes (corresponding to 𝑚 = 2) of the diffraction pattern. Thus,

0.04 𝑚𝑚 (

4 𝑚𝑚 2 )

2 ∗ 1.0 𝑚 = 𝜆 𝜆 = 40 𝑛𝑚

Note: This wavelength is not within the visible

range of light (just placed random values).

𝑎𝑦𝑚,𝑑𝑖𝑓𝑓

THE SPEED OF LIGHT

 Light slows down when travelling in a medium other than air/vacuum

 The ratio between the speed of light in vacuum (c) and its speed in some medium (v) is given by 𝑛 = 𝑐

𝑣, 𝑐 = 3 × 10

8 𝑚

𝑠

n is called the index of refraction

LAWS OF REFLECTION AND REFRACTION

 Law of reflection:

𝜃₁ = 𝜃₁′  Law of refraction (Snell’s Law)

1

𝑣₁ sin 𝜃1 = 1

𝑣₂ sin 𝜃2 𝑛₁sin 𝜃₁ = 𝑛₂sin 𝜃₂  Total internal reflection

𝜃_𝑐 = sin−1 𝑛2 𝑛₁

(Special case of Snell’s Law where 𝜃₂ = 90°; no light is refracted at angles greater than 𝜃_𝑐)

Note: Law of reflection holds for all types of mirrors, i.e plane and spherical mirrors.

REFLECTION IN SPHERICAL MIRRORS

CYLINDRICAL LENS

𝜃_𝑖 𝜃_𝑟 𝜃_𝑟’ 𝜃_𝑟’ 𝜃_𝑖 𝜃_𝑟

𝜃

_𝑖

: angle of incidence

𝜃

_𝑟

: angle of reflection

𝜃

_𝑟

’: angle of refraction

Glass Air Glass Air

10) AT WHICH OF THE GIVEN SET-UPS DOES REFRACTION OCCUR

AT THE AIR-TO-GLASS INTERFACE?

10) AT WHICH OF THE GIVEN SET-UPS DOES REFRACTION OCCUR

AT THE AIR-TO-GLASS INTERFACE?

Answer:

Hint: Draw a normal line at the air-to-glass interface.

POLARIZATION – MALUS’ LAW

E_o: amplitude of the incident electric field

θ: angle between the polarization of the incident light and the transmission axis

I_trans: intensity of transmitted light

𝐸

_{𝑡𝑟𝑎𝑛𝑠}

= 𝐸

_𝑜

cos(𝜃)

11) GIVEN THE FFG. CONFIGURATION, WHERE SHOULD A THIRD POLARIZER BE PLACED FOR

THE LIGHT SENSOR TO HAVE AN INTENSITY READING GREATER THAN 0 LUX?

12) WHAT ARE THE ALLOWED ANGLES?

𝜃 = 00 𝜃 = 900

A B C

Answers:

B. In between the two polarizers.

Allowed angles: 0

0

_{< 𝜃 < 90}

0

sensor 0 lux

CONCEPTS



Laser diode – linearly polarized light



Plain light source – not linearly polarized

Intensity of transmitted laser diode light source changes as polarizing angle is varied



Malus’ Law

CONCEPTS



Intensity of light source does not affect Malus’ Law behavior.



Both laser diode and plain light source plots exhibit Malus’ Law as seen on

𝐼

_𝑒𝑥𝑝

/𝐼

_𝑚𝑎𝑥

vs. 𝜃 plots



Transverse nature of EM waves is shown with the reduction of intensity

as angle of polarization changes.



No more light is transmitted when the angle of polarization is completely

THIN LENS EQUATION

http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lensdet.html

Positive Negative

object distance

real object in front of the lens (incident side)

virtual object at the back of the lens (transmission side) image distance

real image at the back of the lens (transmission side)

virtual image in front of the lens (incident side) focal length converging/convex lens diverging/concave lens Parameters: 𝑠_𝑜 − 𝑜𝑏𝑗𝑒𝑐𝑡 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑠_𝑖 − 𝑖𝑚𝑎𝑔𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓 − 𝑓𝑜𝑐𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ ℎ_𝑜 − 𝑜𝑏𝑗𝑒𝑐𝑡 ℎ𝑒𝑖𝑔ℎ𝑡 ℎ_𝑜 − 𝑖𝑚𝑎𝑔𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 Lens equation: 1 𝑠_𝑜

+

1 𝑠_𝑖

=

1 f Linear magnification:

𝑀 = −

𝑠𝑖 𝑠_𝑜

=

ℎ_𝑖 ℎ_𝑜

Criteria for formed image:

|M| > 1 : magnified |M| < 1 : reduced |M| = 1 : same height +M : upright

COMBINATION OF TWO LENSES (IN CONTACT)

Effective focal length (in contact):

Concave lens (diverging): negative focal length (-f) Convex lens (converging): positive focal length (+f)

COMBINATION OF TWO LENSES (NOT IN CONTACT)



13) GIVEN THE FOLLOWING PARAMETERS, WHAT IS THE

MAGNIFICATION OF THE IMAGE?

o = 5 cm

f = 4 cm

Answer: 1 𝑜 + 1 𝑖 = 1 𝑓 1 5 𝑐𝑚 + 1 𝑖 = 1 4 𝑐𝑚 𝑖 = 20 𝑐𝑚 𝑀 = − 𝑖 𝑜 = 20 5 𝑀 = 4