Calculate the electric flux passes through the object.A= (Πr2)E= 100 N/CR= 150 cmΦ = 40̊ 2.1. Calculate the electric flux passes through the object.A= (Nr²)E= 100 N/CR= 150 cmO = 40SOLUTION:

The electric flux is defined as ϕ = E A cosθ where, E is the electric field, A is the area, θ is…

Answered: Calculate the electric flux passes…

Science

Physics

Calculate the electric flux passes through the object. A= (Nr) E= 100 N/C R= 150 cm O = 40

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter6: Gauss's Law

Section: Chapter Questions

Problem 2CQ: What are the maximum and minimum values of the flux in the preceding question?

See similar textbooks

Related questions

Q: A cylinder is placed with its long side parallel to the electric fields from a vertical circle. The…

A: Here, a figure is given, showing a cylinder. Magnitude of electric field =60 N/C Radius of flat ends…

Q: What happens with the flux through a surface of fixed area when the surface is tilted in such a way…

A: What happens with the flux through a surface of fixed area when the surface is tilted in such a way…

Q: Find the electric flux (in units of N.m/C) through the surface in the figure ( dashed line) given…

Q: A 92fC is placed inside a cube. What is the amount of electric flux passing through each side

A: The electric flux through the each side of the cube is ϕ =q6εo

Q: The closed triiangle box in the figure is located in the horizontal electric field with the…

A: If a constant/uniform extermal electric field enter and exist from closed surfaces than total flux…

Q: A 3.65x10-12 C is placed inside a cube what is the amount of electric flux passing through each side

Q: Calculate the electric flux passes through the object. A= (Πr2) E= 100 N/C R= 150 cm Φ = 40̊

A: Given data: A= (Πr2)E= 100 N/CR= 150 cmΦ = 40̊ Need to determine the electric flux passes through…

Q: SA -29 S2 -29 S3 S6 39

A: Different charges are placed inside different surfaces as shown in the figure. The charges are : -2q…

Q: A charge emanates an electric field of 112 N/C. If this field is placed on a circular surface with a…

A: Electric field (E) = 112 NCradius (r) = 38 μm angle made by electric field with the surface (θ) = 84…

Q: The flux ϕE=35N⋅m2/C through the open surface A⃗ =8i^+8j^+10k^ is given. Find the value of a if the…

A: Electric flux ϕ= E. A

Q: Find the total electric flux over the horizontal cylinder of base area A. The electric field is…

Q: Which of the surfaces show in the diagram has the largest amount of electric flux passing through…

Q: 2 m 100 N/C 2 m The electric flux through the shaded surface is

Q: In the figure shown, the magnitude of the uniform electric field is E = 70 kN/C . Calculate the…

A: Given that: E=70 kN/C

Q: The electric flux through the shaded surface is А. 0 В. 200 N m/C 2 m С. 400 N m²/C D. Flux isn't…

A: Calculate the cross-sectional area of the shaded surface.

Q: The electric field vector in a certain region is given by E = 2yi + 5xk N/C. Find the electric flux…

Q: q2 q4 +. q1 $4 q3 s3 s2 s1 +

A: Gauss Law states that the net electric flux out of a closed surface is equal to the net charge…

Q: Q3: An insulating solid sphere of radius (10 cm) has a uniform volume a positive charge density…

Q: A 45.0-cm-diameter circular loop is rotated in a uniform electric field until the position of…

A: Given that : d=45 cm =45 x 10-2 m ϕ=5.60 x 105 m2/C

Q: Find the electric flux (in units of N.m2/C) through the surface in the figure ( dashed line) given…

Q: Calculate the flux through a square plate 1.0 m on a side inclined at 120° with respect to a…

A: Given that, Electric field = 100 N/C Inclined angle = 120° Side of plate = 1.0 m

Q: The electric field vector in a certain region is given by E = 2yi + 5xkN/C. Find the electric flux…

A: The electric flux is given by,

Q: A box that is in triangular shape is kept in an electric field of magnitude E = 2 × 103 N /C-1 as…

A: The electric flux is given by ϕ = E A cosθ where, E is the electric field vector , A is the area…

Q: The electric field between two parallel plates has magnitude of 900 N/C and is directed downward.…

A: The diameter d=25cm The magnitude =900N/c

Q: The total electric flux from a cubical box 36.0 cm on a side is 1270 N-m2/C.

A: Let the total flux is from the all surfaces of cube. From Gauss's law, Flux, φ=∫E·dA=qε0

Q: Q3: An insulating solid sphere of radius (10 cm) has a uniform volume a positive charge density 2…

A: According to Gauss's law, the electric flux through a closed surface in a vacuum is given as 1ε0…

Q: A flat sheet is in the shape of a rectangle with sides of lengths 0.400 m and 0.600 m. The sheet is…

Q: 70 2 m 20° 100 N/C The electric flux through the shaded surface is

A: Solution: Given that electric field E = 100 N/C A = 2x2 m2 angle = 90-(70+20) =0 degree

Q: The flux coming out from a circular balloon with a radius of 1.0 x 10 ^-1 m is 1.0 x 10^3 Nm^2/C. If…

Q: A 44.0-cm-diameter circular loop is rotated in a uniform electric field until the position of…

A: We know, Electric flux is given as, ϕ=E→.A→=EAcosθ where θ is the angle between E and A Now flux…

Q: Calculate the electric flux for each of the closed surfaces a, b, c, and d. d. a 93 97 q4 95 98

Q: The total electric flux from a cubical box 31.0 cm on a side is 3890 N-m²/C.

A: electric flux =EA

Q: Calculate the electric flux in each surface: S2 10.0 µC +3.0 µC +1.0 uC +7.0 µC -8.0 µC

Q: The electric ﬂux through a square-shaped area of side 5 cm near a large charged sheet is found to be…

A: Given: Side a=5 cm=0.05 m Electric flux ϕ= 3*10^-5 Nm2/C To find the charge density on the sheet.

Q: Determine the electric flux through each surface whose cross-section is shown below. (Use the…

A: CONCEPT:

Q: q2 q4 + q1 $4 q3 + s3 s2 s1

A: Given value--- q1=q2=q3=q4=q5=150nC . We have to find--- What is the electric flux of s1?

Q: a radius of a and is placed in a uniform electric field oriented at an angle of 45° from the…

Q: An electric field of 800N/C makes an angle of 50.00 with the surface vector, which has a magnitude…

Q: The electric field between two parallel plates has magnitude of 900 N/C and is directed downward.…

A: The magnitude =900N/c The diameter =25cm

Q: electric field of 500 V/m makes an angle of 30.00 with the surface vector. It has a magnitude of…

Q: What is the electric flux for the following sets of variables? E= 8.2 x 1013…

Q: A box with a cross-section of 1m x 1m x 1m is surrounding a plan of charge. The uniform electric…

A: Given that---- dimension of box = 1m×1m×1m Uniform electric field = 8N/C Question ❓ Total electric…

Q: Activity 2: SOLVE THE SHAPE OF YOU! Directions: Solve for the value of Area and Electric flux.…

Q: 19. What will happen to the electric flux if the electric field is parallel with the normal line?…

A: Flux = E.A = EAcosϴ , where ϴ is the angle between electric field and the area vector (normal…

Q: An electric field of intensity 2.70 kN/C is applied along the x-axis. Calculate the electric flux…

A: For a uniform electric field passing through a plane surface,ϕE=E→.A→=EAcosθ whereθ is the angle…

Q: In the figure shown, the magnitude of the uniform electric field is E = 70 KNIC . Calculate the…

A: Electric flux:- It is the number of electric field line passes through any surface is called as…

Q: Determine the value of the electric field in a disk with a radius of 12 cm and tilted at an angle of…

Q: The flux associated with a spherical surface is 10 N m2C1. What will be the flux, if the radius of…

Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

Calculate the electric flux passes through the object.
A= (Πr2)
E= 100 N/C
R= 150 cm
Φ = 40̊

2.
1. Calculate the electric flux passes through the object.
A= (Nr²)
E= 100 N/C
R= 150 cm
O = 40
SOLUTION:

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

How far from a charged sphere of 5.4 x 10-9 C will the electric field intensity be 2160 N/C?
A 3.65x10-12 C is placed inside a cube. What is the amount of electric flux passing through each side? Group of answer choices 4.12x10^-1 Nm^2/C 4.12x10^1 Nm^2/C 4.12x10^-1 Nm^2/C^2 4.12x10^1 Nm^2/C^2
What is the net electric flux in S3?
The rectangular surface in the figure below is in a uniform electricfield region with strength of 45.4 N/C. If the surface has a dimensionof 5.00 m × 9.00 m, and the electric flux through the surface is 1040N · m2/C^2, the angle is?
What is the electric flux of s1? q1=q2=q3=q4=q5=150nC
A thin ring whose center is 35cm away has a radius of 52cm. If an electric field of 43700 N/C is measured how much charge is on the ring
What is the electric flux of a sphere with a 1.5m radius experiencing a 102.3 N/C at a 38° angle above the surface of the sphere? Group of answer choices 1.78x10^3 Nm^2/C 1.78x10^-3 Nm^2/C 1.78x10^3 Nm^2/C^2 1.78x10^-3 Nm^2/C^2
Consider a triangle ABC with sides AB = 14.9 cm and BC = 9.7 cm. Angle ABC is 38.4 degrees. What is the length of AC?
What are the maximum and minimum values of the flux in the preceding question?
A 47.0-cm-diameter circular loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be 5.94 105 N · m2/C. What is the magnitude of the electric field?
3 A circular disc with an area of 10m^2 is subjected to an electric field of 5N/C. The electric field creates a 15° angle with a line drawn perpendicular to the surface of the disc. What is the value of the electric flux? 12.94 Nm^2/C 6.47 Nm^2/C 48.30 Nm^2/C 1517.27 Nm^2/C
The electric flux through a cubical box 26 cm on a side is 9.6×10^3 N⋅m2/C. What is the total charge enclosed by the box? Hint The total charge enclosed by the box is μC.