Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 24, Problem 24.49AP
Find the electric flux through the plane surface shown in Figure P23.37 if θ = 60.0°, E = 350 N/C, and d = 5.00 cm. The electric field is uniform over the entire area of the surface.
Figure P23.37
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A uniform electric field of magnitude E = 435 N/C makes an angle of O = 65.0° with a plane surface of area A = 3.50 m2 as in Figure P15.40. Find the electric flux through this surface.
Consider the uniform electric field E = (3.5 j + 3.5 k) × 103 N/C.
a) Calculate the electric flux through a circular area of radius 1.75 m that lies in the yz-plane. Give your answer in N·m2/C.
b) Repeat the electric flux calculation for the circular area for the case when its area vector is directed at 45° above the xy-plane. Give your answer in N·m2/C.
A uniform electric field of magnitude E = 435 N/C makes an angle of θ = 65.0° with a plane surface of area A = 3.50 m2 as in Figure P15.44. Find the electric flux through this surface.
Chapter 24 Solutions
Physics for Scientists and Engineers, Technology Update (No access codes included)
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- A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux through the loop is a maximum. At this position, the electric flux is 7.50 x 105 N-m²/C. Determine the magnitude of the electric field. O 2.44 × 106 N/C O 4.24 × 106 N/C O 1.07 x 106 N/C O 8.88 x 105 N/C O.6.00 x 106 N/C eTextbook and Mediaarrow_forwardFigure P15.49 shows a closed cylinder with cross-sectional area A = 2.00 m². The con- stant electric field É has mag- nitude 3.50 x 10% N/C and is directed vertically upward, perpendicular to the cylinder's top and bottom surfaces so that no field lines pass through the curved surface. Calculate the electric flux through the cylinder's (a) top and (b) bottom surfaces. (c) Determine the amount of charge inside the cylinder. Figure P15.49arrow_forwardA 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/Carrow_forward
- a circular surface with a radius of 0.061 m is exposed to a uniform electric field of magnitude 1.88E4 N/C. The electric flux through the surface is 77 Nm^2/C. What is the angle between the diraction of the electric field and the normal to the surfacearrow_forwardConsider the uniform electric field E = (3.5 jˆ + 3.5 kˆ) × 103 N/C. A) Calculate the electric flux through a circular area of radius 1.75 m that lies in the yz-plane. Give your answer in N·m2/C. B) Repeat the electric flux calculation for the circular area for the case when its area vector is directed at 45° above the x-axis (pointing in the direction of iˆ+kˆ). Give your answer in N·m2/C.arrow_forwardA circular surface with a radius of 0.066 m is exposed to a uniform electric field of magnitude 1.90 × 104 N/C. The electric flux through the surface is 52 N·m2/C. What is the angle between the direction of the electric field and the normal to the surface?arrow_forward
- A constant electric field passes through a plane surface shown in the figure. E d a. If d=15 cm, 0=70.0 ° and the net flux through the surface is 6.00 Nm^2/C, determine the magnitude of the electric field. b. If d=5 cm, 0=60.0 ° and E=350 N/C, determine the electric flux through the surface.arrow_forwardA 3.2-cm-diameter circle lies in the yz-plane with unit vector n^ pointing in the +x-direction. What is the electric flux through the circle if the electric field is E =(2500i^−1500j^+4000k^)N/C?arrow_forwardA cylindrical metal can have a height of 27 cm and a radius of 11 cm. The electric field is directed outward along the entire surface of the can (including the top and bottom), with a uniform magnitude of 4.0x10^5 N/C. a.) What is the surface area of the cylinder?arrow_forward
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