The figure below shows a uniform electric field of magnitude E = 455 N/C making an angle of o = 64.0° with a flat surface of area A = 3.10 m2. what is the magnitude of the electric flux through this surface (in N · m2/C)? E

The figure below shows a uniform electric field of magnitude E = 455 N/C making an angle of o = 64.0° with a flat surface of area A = 3.10 m2. what is the magnitude of the electric flux through this surface (in N · m2/C)? E

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter24: Gauss’s Law

Section: Chapter Questions

Problem 24.48AP: Consider a plane surface in a uniform electric field as in Figure P24.48, where d = 15.0 cm and =...

See similar textbooks

Related questions

Q: The electric flux through a cubical box 2.9 cm on a side is 8.0 X 10³ N. C m²/C. What is the total…

A: Write the expression for the electric flux. Here, qenc is the enclosed charge and εo is the…

Q: 6.2. Figure 4 shows an edge-on view of two planar surfaces that intersect and are mutually…

Q: An electric field of 600N/C makes an angle of 20.00 with the surface vector, which has a magnitude…

Q: Consider the uniform electric field E = (6.0j + 2.0 k) X 103 N/C. What is its electric flux (in N.…

Q: flat surface of area 3.60 m2 is rotated in a uniform electric field of magnitude E = 5.65 105 N/C.…

A: Given, E = 5.65×10⁵ N/C A = 3.60 m²

Q: The electric flux through a cubical box 8.0 cm on a side is 1.2 × 103 N · m2 /C. What is the total…

A: Electric flux

Q: Charges q1=−9Q and q2=+2Q are located at x=−a and x=+a, respectively. a. What is the net electric…

A: Charge q1 = -9Q is placed at x = -a and charge q2 = +2Q is placed at x = +a.Electric Flux through a…

Q: Find the electric flux through the plane surface shown in the figure below if = 61.8°, E = 372 N/C,…

A: We have a plane square surface of side d=4.7 cm=0.047 m through which an electric field of magnitude…

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

Q: A flat circle of radius 19 cmcm is placed in a uniform electric field of magnitude 7.1×102 N/CN/C .…

A: GivenRadius r = 19 x 10-2 mField E = 7.1 x 102 N/C

Q: A flat surface of area 3.50 m² is rotated in a uniform electric field of magnitude E = 5.70 × 105…

A: Given Data: The area of a flat surface is, A=3.50 m2 The magnitude of the uniform electric field is,…

Q: 15.7c Consider a closed triangular box resting within a horizontal electric field of magnitude E =…

Q: The surface shown is a square and has a side length of 11.0 cm. The electric field acting on this…

A: It is given that

Q: A uniform electric field measured over a square surface with side length d = 16.0 cm makes an angle…

A: We know electric fluxϕ=E·A=EAcosθ ----(1) where E= electric field A = area of surface θ=angle…

Q: An electric field of magnitude 2.86 kN/C is applied along the x axis. Calculate the electric flux…

A: Given information: The magnitude of the electric field, E=2.86 kN/C This field is acting along the…

Q: disk of radius 0.10 m is oriented with its normal vector n at 30 degrees to a unform electric field…

Q: Find the electric flux (in units of N.m2/C) through the surface of the rectangle ( 30 cm x 40 cm) in…

A: Given: The electric filed is E=443.5 N/C. The angle is θ=30°. The area is A=30 cm×40 cm. The…

Q: Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.60 x…

Q: What is the area of a rectangle whose electric flux is 3.35x10-10 C2/Nm2 and electric field of 250…

A: The electric flux is ϕ=3.35×10-10 Nm2/C The electric field is E=250 N/C

Q: An electric flux of 144 N.m2/C passes through a flat horizontal surface that has an area of 0.72 m2.…

A: Electric flux through a surface (∅) = 144 N.m2 mCArea of surface (A) =0.72 m2Angle made by electric…

Q: The cube in the figure (Figure 1)has sides of length L=10.0cm. The electric field is uniform, has a…

A: (a) The expression for electric flux through the side 1 is, Substitute the values, to find the…

Q: sphere radius of 13.85 cm creates an electric field of strength of 9.75 N/C at a distance of 21.6 cm…

Q: What is the total flux passing through a 10 cm x 6 cm surface in a region where the electric flux…

A: GIven data: The area of the surface is A=10 cm×6 cm=60 cm2. The electric flux density is E=2700…

Q: An electric field, E = 477 N/C, passes through a circular loop, of area A = 0.485 m², at an angle of…

A: CONCEPT:

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 uniform electric field with a magnitude of 1.25x105 N/C passes through a rectangle with sides of…

A: Given Data: The uniform electric field is E = 1.25 x 105 N/C. The sides of the rectangle are L =…

Q: The dimensions of a cone are displayed in the figure. As shown, the bottom face of the cone rests on…

A: Given - electric field, E = 732 N/C diameter of cone , D = 4.22 cm so , radius of cone…

Q: What is the radius of a flat circle when it is placed in a uniform electric field magnitude of 4.6 x…

Q: A disk of radius 132 mm is oriented with its normal unit vector at 30º to a uniform electric field…

Q: 1. The figure shows a cube of side length a = 12 cm placed in a uniform electric field Ē = 3î – 6ĵ +…

A: 1. Side of cube is a = 12 cm => a = 12 × 10-² m Area A = a² = 144×10-⁴ m² E = 3 i - 6j +4k…

Q: An electric field of magnitude 3.40 kN/C is applied along the x axis. Calculate the electric flux…

A: Electric Flux

Q: 4. A uniform electric field of magnitude 19000 N/C is perpendicular to a square sheet with sides 0.7…

A: Given E= 19000N/C Side of square= 0.7m So A= 0.49m2 Electric Flux= E.A Cosx here E is electric field…

Q: cube with edges of 1.4 m presents the orientation indicated in the figure within a region of a…

Q: What is the area of a rectangle whose electric flux is 3.35x10-10 C2/Nm2 and electric field of 250…

A: Given value--- electric flux = 3.35 × 10-10 C2/Nm2 . electric field = 250 N/C. We have to find---…

Q: What is the magnitude of the electric flux passing through the flat surface, tilted at 75° with…

Q: Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.60 x…

A: Given : E = 7.60 x 104 N/C

Q: Find the electric flux ( in units of N.m2/C) through the surface of the rectangle (30 cm x 40 cm) in…

A: Write the given values of the problem- Electric field(E) E=224.8 N/C Suraface of the rectangle=30…

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

A: The height of the vertical rectangular surface can be obtained as Now, the electric flux through…

Q: The Electric Flux and Gauss’ Law A flat sheet of paper of area 0.250 m2 is oriented so that the…

A: Given data *The given area of the flat sheet is a = 0.250 m2 *The given angle is θ=600 *The given…

Q: A electric field at 3.8 cm from the centre of a long copper rod of radius 0.7 cm has a magnitude 3.6…

Q: An electric field of magnitude 3.42 kN/C is applied along the x axis. Calculate the electric flux…

Q: The rectangular surface in the figure below is in a uniform electric field region with strength of…

Q: What is the radius of a flat circle when it is placed in a uniform electric field magnitude of 4.6 x…

A: Given data: Magnitude of electric field (E) = 4.6 x 102N/C Angle = 45 degree Electric flus ϕ=74.49…

Q: 3. GAUSS'S LAW A uniform electric field with a magnitude of 1.25x10° N/C passes through a rectangle…

Q: The figure shows a rectangular flat surface with a = 1.1 m and b = 1.4 m. A uniform electric field…

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

The figure below shows a uniform electric field of magnitude E = 455 N/C making an angle of o = 64.0° with a flat surface of area A = 3.10 m2. What is the magnitude of the electric flux
through this surface (in N · m2/C)?

Expert Solution

Step 1

Flux Formula

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

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

The nonuniform charge density of a solid insulating sphere of radius R is given by = cr2 (r R), where c is a positive constant and r is the radial distance from the center of the sphere. For a spherical shell of radius r and thickness dr, the volume element dV = 4r2dr. a. What is the magnitude of the electric field outside the sphere (r R)? b. What is the magnitude of the electric field inside the sphere (r R)?
Consider a plane surface in a uniform electric field as in Figure P24.48, where d = 15.0 cm and = 70.0. If the net flux through the surface is 6.00 N find the magnitude of the electric field. Figure P24.48
The figure below shows a uniform electric field of magnitude E = 470 N/C making an angle of ? = 61.0° with a flat surface of area A = 3.10 m2. What is the magnitude of the electric flux through this surface (in N · m2/C)?
A uniform electric field with a magnitude of 1.25x105 N/C passes through a rectangle with sides of 2.50 m and 5.0 m . If the electric flux through the rectangle is 6.6x105 Nm2/C , find the angle between the electric field vector and the vector normal to the rectangular surface ?
Consider a plane surface in a uniform electric field as shown in the figure, where d =20 cm and the angle= 70. The net flux through the surface is 7 N.m2/C. What is the magnitude of the electric field? In N/C.
We have an electric flux of magnitude 141 Nm2/ C passing through a flat horizontal surface with an area of 0.58 m2. The flux is due to a uniform electric field, which points 14.4 degrees above the horizontal. Find the magnitude of the electric field.
A uniform electric field with a magnitude of 1.25x10^5 N/C passes through arectangle with sides of 2.50 m and 5.0 m. If the electric flux through the rectangle is 6.6x10^5 Nm2/C, find the angle between the electric field vector and the vector normal to the rectangular surface?
A circular surface with a radius of 0.062 m is exposed to a uniform electric field of magnitude 1.58 × 104 N/C. The electric flux through the surface is 68 N·m2/C. What is the angle between the direction of the electric field and the normal to the surface?
A uniform electric field with a magnitude of 1.25x105 N/C passes through a rectangle with sides of 2.50 m and 5.0 m . If the electric flux through the rectangle is 6.6x105 Nm2/C , the vector normal to the rectangular surface ?
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?
A uniform electric field of strength E = 334 N/C passes through a flat surface at an angle of 0.880 radians. The electric flux is 455 (N/C)·m2. What is the area of the surface?
A flat surface with an area of 1.25 m² is rotated through a uniform horizontal electric field of 5 N/C. What is the electric flux if the normal of the surface is (a.) perpendicular to the electric field, (b.) parallel to the electric field and (c.) forms a 32° between the electric field?