TRUE OR FALSE 1. When the electric field and the normal line of the surface area are parallel, we can obtain maximum flux2. The angle theta would be zero if the surface area is perpendicular to the electric field. 3. The lines of force passing through a surface of a given area are dependent on the strength of the E-field.4. The Electric field and the amount of electric flux are proportional to one another. 5. Zero Flux will be calculated when the surface area itself is placed parallel to the direction of the electric field.6. Zero flux will be obtained when the normal line of the surface area is perpendicular to the direction of the electric field7. The electric flux is dependent on the orientation of the surface area.8. When the normal line of the surface area is perpendicular to the direction of the electric field the angle theta would be 90°.9. Minimum flux will be obtained when the normal line of the surface area isperpendicular to the direction of the electric field. 10. The unit for electric flux is similar to withthe electric field.​

1. TRUE OR FALSE 1. When the electric field and the normal line of the surface area are parallel, we can obtain maximum flux2. The angle theta would be zero if the surface area is perpendicular to the electric field. 3. The lines of force passing through a surface of a given area are dependent on the strength of the E-field.4. The Electric field and the amount of electric flux are proportional to one another. 5. Zero Flux will be calculated when the surface area itself is placed parallel to the direction of the electric field.6. Zero flux will be obtained when the normal line of the surface area is perpendicular to the direction of the electric field7. The electric flux is dependent on the orientation of the surface area.8. When the normal line of the surface area is perpendicular to the direction of the electric field the angle theta would be 90°.9. Minimum flux will be obtained when the normal line of the surface area isperpendicular to the direction of the electric field. 10. The unit for electric flux is similar to withthe electric field.​

Answer:

1.T

2.t

3.t

4.m

5.t

6.t

7.m

8.m

9.t

10.t

2. Describe an example in which the magnetic flux through a surface is zero and the magnetic field is not zero. ​

What is magnetic flux?

It is a measure of the strength of a magnetic field in a given area. It is also the amount of magnetic in a unit area upright to the direction of magnetic flow.

The magnetic flux will become zero if the surface of it was closed.

Here's an example;

Magnetic flux definition

Flux through a closed surface S.              [tex]\int\limits^s_s[/tex]B.dA=0

This is like Gaúss' law, but for magnetism. However, it states that:

The number of magnetic field lines that enter a volume enclosed by a surface S must equal the number that leave the volume.It implies that magnetic monopoles ( a single electric charge or magnetic pole ) do not exist.

Therefore, the work done by magnetic force which is always equal to zero.

How will you distinguish magnets  from non-magnetized magnetic materials?

For more info. please see related link below.

https://brainly.ph/question/426746

For more info. of magnetic field please see other links below:

https://brainly.ph/question/428587

https://brainly.ph/question/1672581

3. Explain how magnetic flux can be zero when the magnetic field is not zero

Answer:

If the magnetic field is parallel to the place of the area exposed

Explanation:

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4. Compare the electric flux through the surface of a cube of side length a that has a charge q at its center to the flux through a spherical surface of radius a with a charge q at its center. ​

Both have the same flux, as the charge enclosed in the gaussian surface is equal (qq), and that flux can be calculated by Gauss’s law:

Hope it's helps

Have a nice day

5. What is the relationship of the net charge enclosed in a box to the net electric flux through the surface of the box?

Answer:

Directly Proportional

Explanation:

The net electric flux through the surface of a box is directly proportional to the magnitude of the net charge enclosed by the box.

#Brainliest Bunch

Answer:

The net electric flux due to the net charge enclosed by a surface is given by (Gauss Theorem)

Explanation:

doum-i doeeossgileul balabnida

6. when does flux melting occur It occurs when a body of rock near melting point​

Answer:

mixes with a substance that lowers its melting temperature.

7. 1. The figure on the right shows electric charges and Gaussian surfaces, S1, S2, S3, S4, and S5. Solve for the electric flux flowing through all five surfaces.​

Answer:

in the picture

Explanation:

Hope it helps

Correct me if I'm wrong

Brianlest please

8. When electric field lines are perpendicular to the surface, electric flux is?

Answer:

Baliktarin mo yung YEHEY oh d ba walang nagbago,

ikaw pa rin gusto ko.

Explanation:

[Just Kidding]

9. If the net electric field is zero then the net electric flux is also zero? Please explain

Answer:

Gauss’s law for electric flux through a closed arbitrarily shaped surface due to charge distribution in it states that:

The net electric flux through a closed surface is [math]1/E_{°} [/math]times the total charge enclosed in it.

Mathematically, [math]ø_{e}=Q / E_{°}[/math]

That means[math] [/math] [math]ø_{e}[/math] is zero only when:

Charge enclosed Q=zero

Or their is Equipotential surface due to equal and opposite charges enclosed in it due to which there is zero potential difference i.e.,[math] ◇V = zero [/math](and hence zero electric flux)

10. 12. Flux reversal is encoded using method calleda. Non-return to two b. non-return to zeroc. non-return to one​

Answer:

B

Explanation:

b.non-return to zero

Answer:

B. Non-return to zero

Explanation:

Hope it helps!

11. What is the total electric flux through a closed surface containing a 2.0 µc charge?

Answer:

sorry need lang ng points hahaya

12. What is the ratio of electric fluxes through the two surfaces? ΦA : ΦB 5 :​

Answer:

+67 correct me if im wrong coz I don't really study alot of these and don't be mean if I'm wrong jeez if I'm wrong just find another. and. answer tho there's alot

13. Which between decreasing the distance between the surface and the light in half, and doubling the luminous flux would increase the illuminance on a surface more? plss

''A small light source with intensity uniform in all directions is mounted at a height of 10 metres above a horizontal surface. Two points A and ..

14. Compare the electric flux through the surface of a cube of side length a that has a charge q at its center to the flux through a spherical surface of radius a with a charge q at its center.

Answer:

Both have the same flux, as the charge enclosed in the gaussian surface is equal (qq), and that flux can be calculated by Gauss’s law..

Explanation:

dunno if correct.

15. when does flux melting occurs​

Answer:

flux melting occurs when water and other volatile components are introduced to hot solid rock, depressing the solidus enough.

16. What is the total electric flux through a closed surface containing a 2.0 charge?

Answer:

The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux through an area is defined as the electric field multiplied by the area of the surface projected in a plane perpendicular to the field.

Answer:

The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux through an area is defined as the electric field multiplied by the area of the surface projected in a plane perpendicular to the field.

17. determine the electric flux for a gaussian surface that contains 100million electron​

see attached image

for the answer

18. 1. Determine the electric flux for a Gaussian surface that contains 50 million electrons,

Answer:

the electric flux through a gaussian surface will be 1.5 Nm² electrons

Explanation:

HOPE ITS HELP

AND MARK ME AS BRAINLIST

19. A point charge of 3.0 µC is at the center of a cubical Gaussian surface 90cm on edge. What is the net electric flux through the surface?​

Calculating the electric flux

Given

Point Charge (Q) = 3.0 µCLength of edge = 90cmPermittivity of Free Space (∈₀) = 8.85 x 10⁻¹² F/mElectric Flux (Φₑ) = ?

According to Gauss Law, the electric flux is always depends to the amount of point charges. The length of a Gaussian surface is always discarded, as also the point chage is enclosed inside the figure.

Φₑ = Q / ∈₀

Φₑ = 3 x 10⁻⁶ C / 8.85 x 10⁻¹² F/m

Φₑ = 3.38 x 10⁵ N • m² / C

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20. If the electric flux of a Gaussian surface is approximately 5.43x10^-6, inward, how many and what type of particle is inside the surface?

Answer:

Lima kaboog kakaletler

21. What is the total electric flux through a closed surface containing a 2. 0 uC charge?.

Answer:

Φ = q ϵ 0 = 2 ⋅ 1 0 − 6 8.85 ⋅ 1 0 − 12 = 2.26 ⋅ 1 0 5 N ⋅ m 2 C

22. A point charge of 3.5 μC is at the center of a cubical surface 70 cm on edge. What is the net electric flux through the surface?

Answer:

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . , . .. .

23. Gauss law if the electric flux through a spherical surface is zero, then there is no charge within the sphere.

Answer:

Demonstrated Knowledge of Process

Student knows and was able to identify and explain necessary theories/ tasks for completion of the project. Relies on own memory skills to get task done.

Followed exact directions to activity given in lap book and by instructor,

Student was able to complete the tasks without assistance.

Student followed all safety rules.

Student was able to identify and explain necessary theories/ tasks for completion of the project with sorme assistance. Uses own words to describe the task.

Followed directions.

Student was able to complete the tasks with little assistance.

Good Student followed most safety rules but have forgotten some.

Student was unable to identify or explain concepts without major prompting. Requires adult assistance to get job done. Uses very little vocabulary to describe the task.

Moderately followed directions. Worked at a pace that was productive.

Student was able to complete the tasks with moderate assistance.

Fair Student attempted to follow safety rules but failed to meet several

Student was not able to both identify and explain major theories/tasks. Uses others views to explain the tasks and doesn't complete any steps on their own for any of steps.

Did not follow directions in any of the tasks and at times refused to slow down to do the tasks well.

Student was unable to complete the tasks without major assistance.

Undeveloped Student failed to follow a significant number of safety rules.

Student lacked interest in demonstrating knowledge of project and/ or process,

Student was non-compliant when given directions all the time.

Student refused or was unable to start the project when offered assistance

Non-Compliant Student was not safe enough to enter/ participate in class

Ability to Follow Directions

Level of Needed Assistance

Application of Safety Practices

24. what will happen to the electric flux of the line is not perpendicular to the surface?​

Answer:

If the surface is perpendicular to the field (left panel), and the field vector is thus parallel to the vector, →A, then the flux through that surface is maximal. If the surface is parallel to the field (right panel), then no field lines cross that surface, and the flux through that surface is zero.

25. Shows that the magnetic flux is maximum when the angle between a magnetic field and normal to an area is zero.

Answer:

The correct option is A True

Φ

= B

×

A

×

(cos

θ

).

Where,

Φ

= Magnetic flux,

B = magnetic field,

A = area of the plane, and

θ

= angle between the normal to the plane and the magnetic field.

The area and the magnetic field are constant. Hence, the magnetic flux will only depend on

θ

.

When

θ

=

0

∘

,

Φ

= BA =

Φ

m

a

x

as cos(

0

∘

) = 1, which is the maximum value the cosine function can attain.

26. 4. As charge increases, what happens to flux density?A. IncreasesB. DecreasesC. Remains constantD. Becomes zero​

Answer:

A foe

Explanation:

basta A foe ang alam ko

27. 1. An electric field of 600 V/m makes an angle of 40° with the surface vector. Thearea is 0. 500 m2. Find the electric flux that passes through the surface.​

Answer:

[tex] \therefore [/tex] The electric flux that passes through the surface is 228Vm

Explanation:

As the problem, given that 600V/m (E) makes an angle of 40 (θ) and it has area of 0.500 (A). Now, using the formula below

[tex] \boxed{\bold{\Phi_E = EA \ cos \theta}} [/tex]

Substitute the given to the formula

[tex] \rm{\Phi_E = (600)(0.500) cos 40} \\ \\ \rm{\Phi_E = 300 cos 40} \\ \\ \rm{\Phi_E = 300 \cdot 0.76} \\ \\ \rm{\Phi_E = 228} [/tex]

[tex] \therefore [/tex] The electric flux that passes through the surface is 228Vm

28. What is the total electric flux through a closed surface containing a 2.0 charge?

[tex]ϕ = \frac{2.0 × {10}^{ - 6} C}{

8.85 x {10}^{ - 12} {c}^{2} /N. m²

.} \\ = 2.3 × {10}^{5} N. m²/C[/tex]

Understanding Gauss's law

Gauss's law was formulated by a mathematician or physicist Carl Friedrich Gauss (1777-1855). In his findings, Gauss formulated a law that determines the magnitude of an electric flux (total electric field lines that penetrate a certain surface area) that penetrates a closed surface is proportional to the total electric charge that surrounds the closed surface.

He succeeded in finding a formulation for the concept of field lines in the form of mathematical equations. Where, Gauss's law can be used to calculate the electric field strength in certain symmetrical cases.

Mathematically Gauss's law can be formulated as follows:

Φ = ∮AB.dA = EAcosϕ = q/ε0

where:

ε0= 8.85 x 10⁻¹² C²/N. m² is the permittivity of free space.

To formulate Gauss's law, we can use the concept of electric flux, the next review is the electric flux for a closed surface. Suppose there is a point charge q, the electric field strength E resulting from point charges enclosed by a closed area that is spherical with radius r with charge q at the center of the ball.

learn more about gauss's law at https://brainly.ph/question/26194877.

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29. Calculate the electric flux through the rectangle of sides 5cm and 10cm kept in the region of a uniform electric field 100 N/C the angle 0 is 60° Suppose 0 becomes zero,what is the electric flux?

Answer:

The electric flux through a surface is given by the formula:

Explanation:

Φ = E * A * cos(θ)

where Φ is the electric flux, E is the magnitude of the electric field, A is the area of the surface, and θ is the angle between the electric field and a normal vector to the surface.

In this problem, we are given:

E = 100 N/C (magnitude of the electric field)

A = (5 cm) * (10 cm) = 50 cm^2 = 0.005 m^2 (area of the rectangle)

θ = 60° (angle between the electric field and a normal vector to the rectangle)

Plugging in the values, we get:

Φ = (100 N/C) * (0.005 m^2) * cos(60°)

= 0.25 N m^2/C

Therefore, the electric flux through the rectangle is 0.25 N m^2/C.

If the angle 0 becomes zero (meaning the electric field is perpendicular to the surface), then cos(0°) = 1 and the formula simplifies to:

Φ = E * A

Plugging in the values, we get:

Φ = (100 N/C) * (0.005 m^2)

= 0.5 N m^2/C

Therefore, if the angle becomes zero, the electric flux through the rectangle will be 0.5 N m^2/C.

#hope it helps
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30. A closed surface encloses a net charge of 10 nc. what is the net electric flux through the surface

Answer:

The net electric flux is zero through any closed surface surrounding a zero net charge.

Explanation:

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