Consider a closed triangular box resting within a horizontal electric field of magnitude E = 6.66 x 10* N/C as shown in the figure below. 10.0 cm 30.0 cm 60.0 (a) Calculate the electric flux through the vertical rectangular surface of the box. kNm²/c (b) Calculate the electric flux through the slanted surface of the box. kl. m²/c (c) Calculate the electric flux through the entire surface of the box. kNm²/C

Consider a closed triangular box resting within a horizontal electric field of magnitude E = 6.66 x 10* N/C as shown in the figure below. 10.0 cm 30.0 cm 60.0 (a) Calculate the electric flux through the vertical rectangular surface of the box. kNm²/c (b) Calculate the electric flux through the slanted surface of the box. kl. m²/c (c) Calculate the electric flux through the entire surface of the box. kNm²/C

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter25: Gauss’s Law

Section: Chapter Questions

Problem 7PQ: A positively charged sphere and a negatively charged sphere are in a sealed container. The only way...

See similar textbooks

Similar questions

A positively charged sphere and a negatively charged sphere are in a sealed container. The only way the charged spheres can be examined is by observing the electric field outside the container. a. Given the depiction of the electric fields in Figure P25.7A, is the net electric flux through the container zero, positive, or negative? Explain your answer. b. Two different spheres are placed inside a container. Given the depiction of the electric fields in Figure P25.7B, is the net electric flux through the container zero, positive, or negative? Explain your answer.
Find the electric flux through the plane surface shown in Figure P23.37 if = 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
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?
The Electric Flux and Gauss’ Law A flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60° to a uniform electric field of magnitude 14 N/C. A) Find the magnitude of the electric flux through the sheet. B) Does the answer to part (A) depend on the shape of the sheet? Why or why not? C) For what angle ϕ between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest?
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)?
The figure shows a gaussian surface with flat side A and semi-spherical side B. The electric flux through side A is 730 N/(Cm2), and the electric flux through side B is -450 N/(Cm2). What is the total charge inside the surface? 1.180×103 C 2.800×102 C 2.478×10-9 C 5.699×10-10 C 1.044×10-8 C
Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.80 X 104 N/C as shown in Figure P24.4. Calculate the electric flux through (a) the vertical rectangular surface, (b) the slanted surface, and (c) the entire surface of the box.
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.
A cylinder with radius r = 0.75m and length l = 0.6m that has an infinite line of positive charge running along its axis. The charge per unit length in the line is 7µc/m. i. What is the electric flux through the cylinder due to this line of charge? ii. What is the flux through the cylinder if its radius increased to r = 0.32m? iii. What is the electric field intensity at radius r = 0.32m and l = 0.6m? iv. What is the flux through the cylinder if its length increased to l=0.8m? v. What is the electric field intensity at radius r= 0.32 and l=0.8m? vi. Briefly explain the effect of an increase or decrease of radius and length on the electric field intensity.
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?
The surface shown is a square and has a side length of 11.0 cm. The electric field acting on this area has a magnitude of 108 N/C at an angle of 29.3°. Calculate the electric flux through the shown surface. -0.640 N m2/C 1.14 N m2/C 1.30×104 N m2/C 1.14×104 N m2/C
Consider a closed triangular box resting within a horizontal electric field of magnitude E = 8.26 104 N/C as shown in the figure below. (a) Calculate the electric flux through the vertical rectangular surface of the box.