Answered: An electric field (E=4.5N/C) passes…

An electric field (E=4.5N/C) passes through a flat surface area (0.30 m) with an angle of 45˚. What will be the magnitude of the electric flux?

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 24.63CP: A dosed surface with dimensions a = b= 0.400 111 and c = 0.600 in is located as shown in Figure...

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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.

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An electric field (E=4.5N/C) passes through a flat surface area (0.30 m)
with an angle of 45˚. What will be the magnitude of the electric flux?

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A dosed surface with dimensions a = b= 0.400 111 and c = 0.600 in is located as shown in Figure 124.63. The left edge of the closed surface is located at position x = a. The electric field throughout the region is non- uniform and is given by E = (3.00 + 2.00x2)i N/C, where x is in meters. (a) Calculate the net electric flux leaving the closed surface. (b) What net charge enclosed by the surface?
An electric flux of 3.45 x 10^6 Nm^2/C was measured through a rectangle with sides of 7.00 m and 5.00 m. The angle between the electric field vector and the area vector to the rectangular plane is 30.0 degrees. What is the electric field that passes through the rectangle?
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An electric flux of 144 N.m2/C passes through a flat horizontal surface that has an area of 0.72 m2. The flux is due to a uniform electric field. What is the magnitude of the electric field if the field points 15o above the horizontal?
A uniform electric field E = 5000 N/C is passing through a flat square area A= 2m2 . Determine the electric flux. θ = 60o (the angle between the electric field direction and a line drawn perpendicular to the area)
The figure below shows a uniform electric field of magnitude E = 410 N/C making an angle of = 66.0° with a flat surface of area A = 3.60 m². What is the magnitude of the electric flux through this surface (in N. m²/C)? E N - m²/c Ⓡ
A sphere radius of 13.85 cm creates an electric field of strength of 9.75 N/C at a distance of 21.6 cm from the center of the sphere. What is the electric flux through the surface of the sphere?
A uniform electric field of magnitude E = 38.5 N/C makes and angle of θ = 67.3° relative to a line normal to a surface with area A = 3.01 m2. What is the electric flux (Φ) that passes through this surface? Your answer should be in N·m2/C
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Question 2: Calculating Flux What is the electric flux through the surface shown in the figures below: (a) 1 (b) 300 E 60⁰ 15 cm x 15 cm 180 N/C //ww 10 cm x 20 cm 130⁰ F 200 N/C (c) The electric flux through the surface shown in the figure below is 25 N m2/C. What is the electric field strength? 10 cm x 10 cm