GENERALPHYSICS 2

2nd Semester - Module 1

ELECTRIC CHARGES

AND ELECTRIC FIELDS

\

12

Republic of the Philippines

Department of Education Regional Office IX, Zamboanga Peninsula

General Physics 2 - Grade 12 (STEM) Support Material for Independent Learning Engagement (SMILE) Module 1: Electric Charges and Electric Fields First Edition, 2021

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1

What I Need to Know

This module will help you in understanding the basic concepts of electric charges

and fields. The topics covered by this module are electric charges. At the end of this

module, you should be able to:

1. Describe using a diagram charging by rubbing and charging by induction

STEM_GP12EM-IIIa-1;

2. Explain the role of electron transfer in electrostatic charging by rubbing

STEM_GP12EMIIIa-2;

3. Describe experiments to show electrostatic charging by induction

STEM_GP12EM-IIIa-3;

4. Calculate the net electric force on a point charge exerted by a system of point

charges STEM_GP12EM-IIIa-6;

5. Describe an electric field as a region in which an electric charge experiences a

force STEM_GP12EM-IIIa-7; and

6. Calculate the electric field due to a system of point charges using Coulomb’s law

and the superposition principle STEM_GP12EM-IIIa-10.

What’s In

In General Physics 1, your journey explored the various fundamental forces found

in nature. Gravity, one of the forces you studied, was examined in a detailed manner

and how it influences the movement of physical bodies.

This time, we will be exploring the electromagnetic force, one of nature’s

fundamental forces, which possesses both electric and magnetic force. However, we

need to know how this interaction involves particles with electric charge in

understanding this force. This could also be fundamentally represented by mass. When

an object with mass is accelerated by an applied force, objects with electric charges are

also accelerated by the presence of electric forces.

This behavior can be observed when we see lightning strikes the sky, feel the shock

from a metallic surface after scrubbing our shoes across a carpet or when lighter objects

stick with other objects such as dust clinging on a plastic paper.

Activity 1: GETTING RECHARGED!

Direction: This is to check what you have learned about electric charges and electric

fields. Circle the letter of the best answer.

1. Two unlike charges_________ A. attract each other C. neutralize each other B. repel each other D. have no effect on each other 2. Which of the following is not a process of charging? A. Induction B. Friction C. Conduction D. Convection 3. Material A is positively charged. When brought near to material B, they attract.

Which of the following is true? A. Material B is negatively charged C. Material B is uncharged B. Material B is positively charged D. Both are uncharged

2

4. Material B has become positively charged after rubbing it with Material A. Which of the following statements is correct?

A. Material B loses protons C. Material A loses proton B. Material B gains electrons D. Material A gains electron 5. What will happen when two unlike charges are brought together? They will _____ A. repel each other C. attract each other B. neutralize each other D. no effect on each other 6. If you comb your hair and the comb becomes positively charged, then your hair becomes _________. A. positively charged C. uncharged B. negatively charged D. discharged

For No. 7 & No. 8 study the given Triboelectric series, where moving up means positive and moving down means negative.

7. Which of the following pairs has the strongest electrical force of attraction?

A. Welcru and Lokfu C. Xatzki and Melgi B. Zysmu and Melgi D. Kharmi and Xatzki 8. Which of the following would have a negative net charge when rubbed with Kharmi? A. Lokfu B. Welcru C. Xatzki D. Melgi

9. A negatively charged rod is brought near a metal can that rests on a wooden box. You touch the opposite side of the can momentarily with your finger. If you remove your finger before removing the rod, what will happen to the can?

A. It will be discharged C. It will become negatively charged B. it will become positively charged D. Its charge will remain as it was 10. Which of the following can be attracted by a positively charged object? A. Another positively charged object C. A neutral object B. Any object D. No other object

What’s New

Activity 2: WATER BENDING Direction: This activity will help you acquire real-life concepts of static electricity. Materials You Need: 3 Styrofoam cups (you can also use 2 paper cups and an inflated balloon) and a toothpick. Also, this experiment will also require water and someone with dry and clean hair.

1. Prepare the set-up by pushing the toothpick at the bottom side of the cups. Leave the toothpick to produce a gentle drop of water after filling the cup. Hold the cup directly over the second cup. Fill the cup (with a toothpick) with water and check if it is leaking steadily.

2. Observe the flow of water from the top cup to the cup below. What are your observations? ____________________________________________________

3. Rub the third cup on someone with dry hair several times. (This process will help you in acquiring electrical charges).

4. Hold the cup (rubbed against dry hair) near the water stream without getting the cup

wet. What happened to the water flow? _______________________________________

_______________________________________________________________Then, slowly move

the cup away from the stream and observe. Did you observe changes? ___________ If

yes, describe the change. _________________________________________________________

Melgi

Xatzki

Lofku

Khamri

Welcru

Zysmu

3

____________________________________________________________________________________

5. Try other objects (such as balloon, paper cup or any material) aside from the Styrofoam cup and rub it on a dry and clear hair. What objects have you tried? _____________________________Which of the objects have changed the flow of water? _______________________________________________________________________________

What Is It

Electric Charges

The main building block of matter is composed of atoms and molecules. Its

properties are primarily influenced by the electrically charged particles – proton,

electron, and neutron. The table below shows the properties of the charged particles in

terms of mass, charge, and location.

Particle Mass

(in terms of kg)

Charge

(in terms of Coulomb

(C))

Location in atom

proton 9.1093897 x 10-31 kg +1.60217733 x 10-19 C nucleus

electron 1.6726231 x 10-27 kg -1.60217733 x 10-19 C outside nucleus

neutron 1.6749286 x 10-27 kg none nucleus

Materials contain a huge amount of positively charged particles called protons and

negatively charged particles called electrons. When there is an equal number of protons

and neutrons in a matter, the body is electrically neutral. In making a body negatively

charged, electrons are added to a body. On the other hand, a positively charged body

removes electrons.

The transfer of electrons from one body to another proves the law of conservation

of charges. These charges are neither created nor destroyed. According to the principle

of conservation of charges, the sum of electric charges of a body within a closed system

is always constant.

In understanding electrostatic interactions, we have to keep in mind the following

conventions:

1. Any charged object can attract a neutral object.

Number of Number of protons = electrons

negatively charged

positively charged

4

2. Unlike charges attract

3. Like charges repel.

Charging Objects

Materials that allow the movement of electrons from one region to another

are called conductors of electricity, while materials that do not allow the flow of electrons

are called insulators. The majority of the metals are conductors, while nonmetals are

insulators. The electrons can move while protons and neutrons are bound to remain

fixed in the positive nuclei.

Charging objects could happen through induction and conduction. These

charging processes can be demonstrated through an electroscope. The electroscope is

composed of a metal knob, metal rod, glass container, and foil (leaf). The small metal

foils are hung at the end of the metal rod. This should freely move since they open after

being charged.

When a charged object is placed near the metal knob, this causes the foil to open

up since they are being repelled by the presence of excess charges. The foils drop down

when the charged object is placed away from the electroscope.

Charging by Conduction

Metal knob

Glass

Leaves (foil)

When a negatively charged rod touches the neutrally charged metal knob, the knob attracts the electrons making the leaves negatively charged.

When a positively charged rod touches the neutrally charged metal knob, the rod attracts the electrons making the leaves positively charged.

5

Charging by Induction

Electric Force

Charles Augustin de Coulomb (1736-1806) used a torsion balance in studying

gravitational interaction. He studied the attractive and repulsive forces between charges.

He found out that the magnitude of force decreases when the distance of separation

between the charges increases. This is shown by Coulomb’s law as expressed in the

equation below:

𝑭 = 𝒌|𝑸𝟏𝑸𝟐|

𝒓𝟐

where k is proportionality constant 9 x 109 𝑁𝑚2

𝐶2 , Q1 and Q2 and point charges expressed

in Coulomb (C), r is the distance of separation of two charges expressed in meter (m),

and F is the electrostatic force between the two charges expressed in Newton (N). In SI

units, k is not usually written but as 1

4𝜋∈0 where ∈0 = 8.854 𝑥 109 𝐶2

𝑁𝑚2. This actually

complicates the formula but could somehow help when you encounter other formulas.

If test charges are placed at some angle with respect to other charges, this

involves computing the x and y components of forces. Recall your previous lessons on

vector resolution and trigonometric identities.

Refer to the diagram below:

When a negatively charged rod is placed near the knob, the charges undergo polarization. Positive

charges are near the knob while negative charges stay away from the knob. This makes the foil open up.

Electrons are repelled to the Earth when you touch the knob or when the ground wire is

connected. This makes the electroscope positively charged.

When a positively charged rod is placed near the knob, the charges undergo polarization. Negative

charges are near the knob while positive charges stay away from the knob. This makes the foil open up.

When you touch the knob with a finger or attach it with ground wire, the electrons from Earth move towards the knob making it negatively charged.

F

Fx

Fy

Ɵ

F

6

Solving for x component Solving for y component

𝐹𝑥 = 𝐹𝑐𝑜𝑠𝜃

∑ 𝐹𝑥 = 𝐹1𝑥 + 𝐹2𝑥 + 𝐹3𝑥 + ⋯ + 𝐹𝑛𝑥

𝐹𝑦 = 𝐹𝑠𝑖𝑛𝜃

∑ 𝐹𝑦 = 𝐹1𝑥 + 𝐹2𝑥 + 𝐹3𝑥 + ⋯ + 𝐹𝑛𝑦

Solving for magnitude of resultant force Solving for the direction of resultant force

𝐹𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 = √(∑ 𝐹𝑥)

2

+ (∑ 𝐹𝑦)

2

𝜃 = (

∑ 𝐹𝑦

∑ 𝐹𝑥

)

The next examples will help you understand the application of Coulomb’s Law.

Please prepare your scientific calculator and notebook.

Example 1:

Two charges lie on positive x-axis. Charge A (2.0 x 10-9 C) is 2.0 cm from the origin and

Charge B is 4.0 cm from the origin. (-3.0 x 10-9 C). What is the total force exerted by

these two charges on Charge C (5.0 x 10-9 C) located at the origin?

A What is/are

given?

QA = 2.0 x 10-9 C at 𝑟𝐴 = 2.0 𝑐𝑚

QB = -3.0 x 10-9 C at 𝑟𝐵 = 4.0 𝑐𝑚

QC = 5.0 x 10-9 C at 𝑟𝐶 = (0,0)

B What is asked? F = ? at Q3

C Are the units

consistent with

the formula?

No, distance of separation, r, given should be converted

from cm to m.

Thus, r1 = 0.02 m and r2 = 0.04 m.

D

How will you

draw the

problem?

E

What strategy

must be

employed?

In solving the total force experienced by QC from QA and

QB, forces must be computed individually: FAonC and FBonC.

The vector sum of these forces will determine the net force

exerted by QA and QB on QC.

F Solution

Solving for FAonC:

𝐹𝐴𝐶 = 𝑘|𝑄𝐴𝑄𝐶|

𝑟𝐴2

= 9 𝑥 109𝑁𝑚2

𝐶2 (|(2 𝑥10−9𝐶)(5 𝑥 10−9 𝐶)|

(0.02 𝑚)2 )

= 9 𝑥 109𝑁𝑚2

𝐶2 (|(1 𝑥10−17𝐶2)|

4 𝑥 10−4𝑚2 )

= 9 𝑥 109𝑁(2.5𝑥 10−14 ) 𝐹𝐴𝐶 = 2.25 𝑥 10−4 𝑁

The direction of this force lies along the

negative x-component since like charges

repel. Thus, - 2.25 𝑥 10−4 𝑁

Solving for FBonC:

𝐹𝐵𝐶 = 𝑘|𝑄𝐵𝑄𝐶|

𝑟𝐵2

= 9 𝑥 109𝑁𝑚2

𝐶2 (|(−3 𝑥10−9𝐶)(5 𝑥 10−9 𝐶)|

(0.04 𝑚)2 )

= 9 𝑥 109𝑁𝑚2

𝐶2 (|(1.5 𝑥10−17𝐶2)|

1.6 𝑥 10−3𝑚2 )

= 9 𝑥 109𝑁(9.375𝑥 10−15 ) 𝐹𝐵𝐶 = 8.44 𝑥 10−5 𝑁

The direction of this force lies along the

positive x-component since unlike

charges attract. Thus, 8.44 𝑥 10−5 𝑁

0.02 m

0..04 m

x

0.01 m 0.02 m 0.03 m 0.04 m

QA QB

QC

7

∑ 𝐹 = 𝐹𝐴𝐶 + 𝐹𝐵𝐶 = (−2.25 𝑥 10−4 𝑁) + (8.44 𝑥 10−5 𝑁) = −1.41 𝑥 10−4𝑁

G What is the conclusion?

Therefore, the magnitude of total force experienced by QC from QA and QB is 1.406 x 10-4 N directed to the left.

Example 2:

Two point charges are located in xy coordinate system. A charge 2.00 𝑥 10−9𝐶 is

located at (0,4.00 cm) and the other charge −3.00 𝑥 10−9𝐶 is located at (3.00 cm,

4.00 cm). If the third charge 5.00 𝑥 10−9𝐶 is placed at origin, find the resultant force

at the third charge.

A What is/are

given?

Q1 = 2.0 x 10-9 C at 𝑟1 = (0,4)

Q2 = -3.0 x 10-9 C at 𝑟2 = (3,4)

Q3 = 5.0 x 10-9 C at 𝑟3 = (0,0)

B What is asked? F = ? at Q3

C Are the units

consistent with

the formula?

No, the distance of separation, r, given should be

converted from cm to m.

Thus, r1 = (0 m, 0.04 m), r2 = (0.03 m, 0.04 m) and

r3 = (0,0 m)

D How would you

draw the

problem?

Locating the individual forces through a diagram

E What strategy

must be

employed?

The forces experienced by Q3 from Q1 and Q2 must be computed

individually. However, x and y components must be determined

first. For instance, F2on3 has x and y components. The x

component will be calculated by multiplying the magnitude of

F2on3 by sinƟ. On the other hand, y component will be calculated

by multiplying the magnitude of F2on3 by cosƟ. For F1on3, x

component is zero since it lies along y component.

QC QB FBon

a = 0.04

(adjacent)

b = 0.03 m (opposite)

Solving for unknown (c=?) using Phytagorean theorem:

Solving for unknown (Ɵ = ?)

using trigonometric identities:

0.05 m (hypotenuse)

36.87°

Ɵ

0.04 m

0.03 m

c = ?

Q3

Q1 Q2

Q3

Q2

F1on3

F2on3

Q1

36.87°

F1on3

F2on3 F2on3

y-component

F2on3 x-component

8

F Solution

(1) Solving for F1on3:

𝐹13 = 𝑘|𝑄1𝑄3|

𝑟12

= 9 𝑥 109𝑁𝑚2

𝐶2 (|(2 𝑥10−9𝐶)(5 𝑥 10−9 𝐶)|

(0.04 𝑚)2 )

= 9 𝑥 109𝑁𝑚2

𝐶2 (|(1 𝑥10−17𝐶2)|

4 𝑥 10−3𝑚2 )

= 9 𝑥 109𝑁(6.25𝑥 10−15 ) 𝐹13 = 5.625 𝑥 10−5 𝑁

The direction of this force lies along the

negative y-component since like charges

repel. Thus, - 5.625 𝑥 10−4 𝑁

(2) Solving for F2on3:

𝐹23 = 𝑘|𝑄2𝑄3|

𝑟12

= 9 𝑥 109𝑁𝑚2

𝐶2 (|(−3 𝑥10−9𝐶)(5 𝑥 10−9 𝐶)|

(0.05 𝑚)2 )

= 9 𝑥 109𝑁𝑚2

𝐶2 (|(1.5 𝑥10−17𝐶2)|

2.5 𝑥 10−3𝑚2 )

= 9 𝑥 109𝑁(6 𝑥 10−15 ) 𝐹23 = 5.4 𝑥 10−5 𝑁

The magnitude of the force from Q2

towards Q3 is 5.4 𝑥 10−5 𝑁

(3) Solving for x, y components 𝐹23𝑥 =

5.4 𝑥 10−5 𝑁(𝑠𝑖𝑛 36.87°) = 3.24 𝑥 10−5 𝑁

𝐹23𝑦 = 5.4 𝑥 10−5 𝑁(𝑐𝑜𝑠 36.87°) = 4.32 𝑥 10−5 𝑁

𝐹13𝑥 = 0𝑁 (since F13 lies along y-axis)

𝐹13𝑦 = − 5.625 𝑥 10−4 𝑁

x component y component

F13 0 𝑁 − 5.625 𝑥 10−4 𝑁

F23 3.24 𝑥 10−5 𝑁 4.32 𝑥 10−5 𝑁

Sum 3.24 𝑥 10−5 𝑁 −1.31 𝑥 10−5 𝑁

(4) Solving for magnitude and direction of

the resultant force

𝐹𝑅 = √(∑ 𝐹𝑥)

2

+ (∑ 𝐹𝑦)

2

= √(3.24 𝑥 10−5 𝑁)2 + (−1.31 𝑥 10−5 𝑁)2 𝐹𝑅 = 3.49 𝑥 10−5 𝑁

𝜃 = (∑ 𝐹𝑦

∑ 𝐹𝑥

) = 𝑡𝑎𝑛−1 (−1.31 𝑥 10

−5 𝑁

3.24 𝑥 10−5 𝑁) = −21.94° 𝑜𝑟 338°

G What is the

conclusion?

Therefore, the magnitude and direction of the resultant

force is 3.49 𝑥 10−5 𝑁 , 338°.

You can verify the magnitude and direction of the resultant force using the

graphical method for vector analysis.

Electric Fields

You were introduced to the behavior of electric charges and how these charges

produce attractive and repulsive forces. Aside from these forces, it also creates an

electric field E. The electric field of charge Q is the space surrounding the charge. It also

exerts a force F on any test charge q placed within that region.

9

The electric field is represented by the equation below:

𝑬 =𝑭

𝒒

where E is the electric field expressed in 𝑁

𝐶, F is the electric force expressed in newton

(N), and q is the charge expressed in coulombs (C). If q is positive, the direction of E is

the direction of F. On the other hand, the force on a negative charge is opposite to the

direction of the E.

We can also calculate E given the magnitude and position of all charges

involved. Since 𝐹 = 𝑘|𝑄1𝑄2|

𝑟2 , we substitute this formula in 𝐸 =𝐹

𝑞 . Thus,

𝑬 = 𝒌𝒒

𝒓𝟐

Like dealing with charge situated at some angles, you can similarly perform the

same using the trigonometric and vector resolutions concepts. Refer to the diagram

below. Suppose we have a negative charge at xy plane. We want to calculate the electric

field at point P:

Solving for x component Solving for y component

𝐸𝑥 = 𝐸𝑐𝑜𝑠𝜃

∑ 𝐸𝑥 = 𝐸1𝑥 + 𝐸2𝑥 + 𝐸3𝑥 + ⋯ + 𝐸𝑛𝑥

𝐸𝑦 = 𝐸𝑠𝑖𝑛𝜃

∑ 𝐸𝑦 = 𝐸1𝑥 + 𝐸2𝑥 + 𝐸3𝑥 + ⋯ + 𝐸𝑛𝑦

Solving for magnitude of the

resultant force

Solving for the direction of the

resultant force

𝐹𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 = √(∑ 𝐸𝑥)

2

+ (∑ 𝐸𝑦)

2

𝜃 = (

∑ 𝐸𝑦

∑ 𝐸𝑥

)

Example 3:

A point charge 𝑞1 = +6.00 𝑥 10−9𝐶 is at the point 𝑥 = 0.800 𝑚, 𝑦 = 0.600 𝑚 and a second

point charge 𝑞2 = −2.00 𝑥 10−9𝐶 is at the point 𝑥 = 0.800 𝑚, 𝑦 = 0 𝑚. Calculate the

magnitude and direction of the resultant electric field at the origin due to these

charges.

E

F

F

E

Ex

Ey

Ɵ

E

P

10

A What is/are

given?

q1 = +6.00.0 x 10-9 C at 𝑟1 = (0.800 𝑚, 0.600 𝑚)

q2 = -2.00.0 x 10-9 C at 𝑟2 = (0.800 𝑚, 0 𝑚)

B What is asked? E = ? at (0,0)

C Are the units

consistent with

the formula?

Yes, given values have correct SI units.

D How would you

draw the

problem?

Review field of

charges

Locating the individual electric fields at (0,0):

E What strategy

must be

employed?

The electric fields experienced in origin (0,0) must be

computed individually. Furthermore, x and y components

must be solved for each E1 and E2. We then calculate the

magnitude and direction of the resultant electric field.

F Solution

(1) Solving for F1on3:

E1 = k|Q1|

r12

= 9 x 109Nm2

C2 (|(6 x10−9C)|

(1 m)2 )

= 9 x 109Nm2

C2 (|(6 x10−9C)|

1m2 )

= 9 x 109N(6x 10−9 )

E1 = 54 N

C (outward direction)

(2) Solving for E2:

𝐸2 = 𝑘|𝑄2|

𝑟22

= 9 𝑥 109𝑁𝑚2

𝐶2 (|(−2 𝑥10−9𝐶)|

(0.8 𝑚)2 )

= 9 𝑥 109𝑁𝑚2

𝐶2 (|(−2 𝑥10−17𝐶)|

0.64𝑚2 )

= 9 𝑥 109𝑁(−3.125 𝑥 10−17 )

𝐹23 = 28.125 𝑁𝑁

𝐶

The electric field is directed towards the

positive x-axis since it is a negative

charge (inward direction of field).

(3) Solving for x, y components E1x =

54N

C(sin 53°) = −43.126

N

C (along –x-axis)

E1y = 54N

C (cos 53°)

= −32.498 N

C (along – y axis)

E2x = 28.125N

C

(4) Solving for magnitude and direction

of the resultant force

𝐹𝑅 = √(∑ 𝐹𝑥)

2

+ (∑ 𝐹𝑦)

2

= √(−15.001 𝑁/𝐶)2 + (−32.498 𝑁/𝐶)2 𝐹𝑅 = 35.79 𝑁/𝐶

Solving for unknown (c=?) using Phytagorean

theorem:

Solving for unknown (Ɵ = ?)

using trigonometric

identities:

Q1

1 m (hypotenuse)

53° Ɵ

0.6 m

c = ?

b = 0.03 m (opposite)

37° a = 0.04

(adjacent) 0.8 m Q2

E2

53° E1

E1y

E1x 53°

11

E2y = 0N

C since it lies along x-axis

x component y component

E1 −43.126 N

C -32.498

N

C

E2 28.125 N

C 0

Sum −15.001 N

C −32.4898

N

C

𝜃 = (∑ 𝐹𝑦

∑ 𝐹𝑥

) = 𝑡𝑎𝑛−1 (−32.498 𝑁/𝐶

−15.001 𝑁/𝐶)

= 65.22°

G What is the

conclusion?

Therefore, the magnitude and direction of the resultant force

is 35.79𝑁

𝐶, 65.22°.

You can verify the magnitude and direction of the resultant force using a

graphical method for vector analysis.

Electric Flux

The relationship between electric charge and electric field was also formulated

alternatively in Gauss’s law (Karl Friedrich Gauss 1777-1855). This is logically

equivalent to Coulomb’s law, but this was easier to use in finding electric field for

symmetrical charge distribution.

Gauss’s law general statement is expressed as

𝜙𝐸 = 𝐸𝐴cosƟ

where is the electric flux 𝑁𝑚2

𝐶, E is the magnitude of electric field expressed in

𝑁

𝐶 , Ɵ is

the angle between the normal line of the surface and electric field lines. and A is the

area of a given surface expressed in m2.

Electric flux refers to the amount of electric field lines penetrating a given

surface. The electric flux is maximum if the electric field lines are perpendicular to the

surface or parallel to the normal line of the surface. The electric flux is zero when the

electric field line is parallel to the surface or perpendicular to the normal line of the

surface. At a given angle, the electric flux is directly proportional to the component of

the electric field lines.

Example 4:

A point charge 𝑞 = 8.00 𝑥 10−9𝐶 is at the center of the cube with sides of length 0.200 m. What is the electric flux through one of the six faces of the cube?

A What is/are

given?

q = +8.00 x 10-9 C ; 𝑠 = 0.200 𝑚

B What is asked? 𝞍 = ? at one of the faces of the cube

12

C Are the units

consistent with

the formula?

Yes, given values have correct SI units.

D How would you

draw the

problem?

Hence, the distance from the charge to the surface of the cub is 0.141 m

E What strategy

must be

employed?

Using Gauss’s law, substitute E with formula for Electric Field

𝑬 = 𝒌𝒒

𝒓𝟐

F Solution

𝐸 = 𝑘𝑞

𝑟2= 9 𝑥 109

𝑁𝑚2

𝐶2(

|(8 𝑥10−9

𝐶)|

(0.141 𝑚)2)

= 9 𝑥 109𝑁𝑚2

𝐶2(

|(8 𝑥10−9𝐶)|

0.0199 𝑚2) = 9 𝑥 109

𝑁𝑚2

𝐶2 (4.02 𝑥 10−7)

𝐸 = 3,621.55 𝑁/𝐶 𝐴 = 𝑠2 = (0.200 𝑚)2 = 0.04 𝑚2

𝜙𝐸 = 𝐸𝐴𝑐𝑜𝑠Ɵ = 3,621.55𝑁

𝐶(0.04 𝑚2) = 144,862

𝑁𝑚2

𝐶

G What is the

conclusion? Therefore, the electric flux at any surface is 144,862

𝑁𝑚2

𝐶

What’s More

Activity 3: QUALITATIVE PROBLEMS

Direction: Solve the following problems as directed. You may show your solution on a

separate sheet of paper

(1) Two spheres of equal mass and equal charge are separated at a distance r. (a)

Derive an expression for the quantity of charge that must be on each sphere so that

the spheres are in an equilibrium where attractive and repulsive forces are balanced.

(b) How would doubling the distance between spheres affect the expression for the

value of q from the previous problem. Explain.

_____________________________________________________________________________

_____________________________________________________________________________

____________________________________________________________________

0.200 m

0.200 m

0.200 m

0.200 m c = 0.283

c = 0.141

13

(2) How would you draw the electric field lines given the test charges below:

(3) Coulomb’s law and Newton’s law of gravitation are similar in structure. Can Gauss’s

law be applied to gravitational fields? _________If so, what changes are

needed?____________________________________________________________________________

__________________________________________________________________________

What I Have Learned

Activity 4: QUANTITATIVE PROBLEMS

Direction: Write your answers on a separate sheet of paper. You may also consult

your Physics teacher.

Three charged spheres are at the positions shown in the figure.

(a) Find the net electrostatic force at sphere B.

(b) Find the net electric field at (4,-3)

Scoring Rubric

Criteria 3 2 1 0

Physics

Approach

The approach is

appropriate and

complete

The approach

contains minor

errors

Some of the

concepts and

principles are

missing or

inappropriate

The solution

doesn’t indicate

an approach

Procedure Mathematical and

logical procedures

are clear,

complete, and

connected

Mathematical

and logical

procedures are

missing/contain

errors

Most of the

mathematical and

logical procedures

All procedures

are incomplete

and contain

errors

Description Diagrams and

symbols used are

appropriate and

complete

Parts of the

diagrams and

symbols contain

errors

Most of the parts of

the diagrams and

symbols are not

useful

The entire

visualization is

wrong or did

not include

visualization.

C

B A

8.2 µC

6.0µC

4.5µC

3 cm

4 cm

14

What I Can Do Activity 5. BUILDING CONCEPT MAP Direction: Create a concept map out from the things you have learned from this module. You can use words, terms, phrases, or formulas in connecting these concepts. Refer to the scoring guide below: Mueller’s Classroom Concept Rubric

Legible (easy to

read)

No (0-1) Yes (2)

Accurate (concepts were used

accurately)

Many inaccuracies (0-2)

A few inaccuracies (3-4)

No inaacuracies (5)

Complete

(sufficient number

of relevant concepts

and relationships)

Limited use of

concepts

(0-2)

Some use of concepts

(3-4)

Sufficient number

of concepts

(5)

Sophisticated (finding meaningful

connections

between relevant

concepts)

Little or none (0-1)

Few meaningful connections

made (2-4)

Some meaningful

connections

made (5-7)

Meaningful and original

insights

demonstrated

(8)

Assessment

Direction: Write the letter of your choice in the space provided. ______1.The diagram below shows the behavior of the electroscope before and after a

positively charged rod is placed near the electroscope knob. This tells us that the movement of electrons is from ______.

a.rod to leaves c.knob to leaves, then back to knob b.leaves to knob d.leaves to knob, then back to leaves _____2.A glass rod was positively charged when rubbed with a silk cloth. The net

positive charge is accumulated because the glass rod a.loses protons c.gains electrons b.loses electrons d.loses electrons _____3.Gravitational forces and electric forces are both a.forces with attractive and repulsive behaviors b.indirectly proportional to the square root of separation between bodies c.directly proportional to the product of the masses and charges d.decreasing when the distance between two bodies is decreasing

15

_____4.At which point is the electric field strength strongest? _____5.Three spheres were brought together. When Spheres A and B are brought

together, they attract. When spheres B and C are brought together, they also repel. Which of the following is true?

a.Spheres A & C have same signs c.Spheres A & C have opposite signs b.Spheres B & C have same signs d.Spheres A & B have similar signs _____6.The first object has a charge of +3nC, and the second object has a charge of 6

nC. Which is true about the electric forces between these objects?

a.F1on2= 2F2on1 b.3F1on2= 6F2on1 c.-6F1on2=2F2on1 d.F1on2=F2on1 For Nos. 7 to 9, refer to the diagram below:

_____7. A positively charged rod was brought near to a metallic plate; what is the type

of charge induced along the side facing the rod? a.positive b.negative c.netural d.depends on number of charges _____8.If the positively charged rod was touched on the neutral metallic plate, what is

the charge of the plate? a.same b.negative c.positive d.depends on number of charges _____9.After the positively charged rod is placed near the plate, a grounded wire was

attached. What is the charge of the plate after the wire is removed? a.negative c.neutral

b.positive d.depends on number of charges _____10.The electric flux through the surface at the right is ____ a.zero c.positive

b.negative d.unknown _____11.A test charge produced an electric field, E, at point 3 m away

from the charge. The point where the field is half its original values is located at a.1.5 m from the charge c.4.5 m from the charge b.6 m from the charge d.at the point of charge _____12.The electric field of a body is directly related to its a.momentum b.kinetic energy c.potential energy d.charge it carries _____13.Charge q1 (26.0 µC) and q2 (-47.0 µC) experienced an electrostatic force of 5.70

N between them at a distance of ____ a.1.39 m b.1.93 m c.2.14 x 10-19 d.1.464 x 10-8

_____14.How many electrons have been removed from a positively charged particle if it has a net charge of 5 x 10-9 C?

a.5 x 10-9 electrons c.1 x 10-8 electrons b.2.5 x 10-9 electrons d.incomplete information

D A C

B

η

16

_____15.A sphere of radius 0.05 m has a charge of 2.0 nC. The charge is said to be

located at the center of the sphere. The magnitude of electric field inside the surface is

a.zero c.thrice the original value b.twice the original value d.half the original value

Additional Activities

Activity 6. SOCIAL CONTEXT Direction: The community is a rich source for learning opportunities in electrostatics. Choose one from the following suggested activities in understanding the importance and utilization of electrostatics in our daily lives:

1. Ask a local weather forecaster/Science teacher/physicist/electrician/engineer through phone calls or interview through Zoom, Google Meet, or Messenger and inquire (with supervision from your parents/guardian/learning facilitator) about lightning, lightning safety, and lightning rods.

2. Conduct simulations on electrostatics, electric fields, Gauss’s law using online sources. From this, write a short reflection. Refer to any sites below:

- https://phet.colorado.edu/en/simulation/charges-and-fields - https://www.falstad.com/emstatic/ - http://web.mit.edu/viz/soft/visualizations/DLIC/doc/simulations/ex

periments/electrostatics/package-summary.html - https://javalab.org/en/category/electricity_en/static_electricity_en/ 3. Suppose you are to demonstrate a trick to kids using electrostatic

charging. How will you demonstrate the trick? List down the materials (preferably available at home), step-step procedures, and brief explanation of how this trick works.

17

Answer Key General Physics 2 Module 1

Activity 3. Qualitative Problems

Activity 4. Quantitative Problems

Activity 1: Getting Recharged 1. A 2. D 3. A 4. A 5. A 6. B 7. D 8. C 9. B 10. C

Activity 2: Water Bending 2. vertical, thin, unbroken stream of water 4. the stream of water was bent towards the cup; yes

(there should be changes); the stream went back to its original flow

5. answers may vary; the influence of charged objects to water bending may vary depending on their ability to gain or lose charges

18

References

Printed Resources

Sears, F., Zemansky, M. and Young, H. (1992). College Physics 7th Edition. Addison-Wesley

Publishing Company

Zitzewits, Haase and Harper (2013). PHYSICS Principles and Problems. The MAcGraw-Hill

Companies, Inc.

Online References

Harmon, K. (2011). Bend water with static electricity. Scientific American.

https://www.scientificamerican.com/article/static-electricity-bring-science-home/

Java Lab (n.d.). Static Electricity Simulation. Retrieved last February 18, 2021 from

https://javalab.org/en/category/electricity_en/static_electricity_en/

Massachusetts Institute of Technology. (n.d.) Package simulations experiments electrostatics.

Retrieved last February 20, 2021 from

http://web.mit.edu/viz/soft/visualizations/DLIC/doc/simulations/experiments/electros

tatics/package-summary.html

Mueller, J. (n.d.) Concept map rubric. Retrieved last February 22, 2021 from

https://teach.its.uiowa.edu/sites/teach.its.uiowa.edu/files/docs/docs/Concept_Map_Ru

brics_ed.pdf

PHET (2021). Charges and Fields. University of Colorado. Retrieved last February 19, 2021

from https://phet.colorado.edu/en/simulation/charges-and-fields

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