Concept explainers
In the following Questions, a Gaussian cylinder with radius a andl is placed in various electric fields. The end caps are labeled A and C and the side surfaces is labeled B. In each case, base your answer about the net flux only on qualitative arguments about the magnitude of the flux through the end caps and side surfaces.
D. A positive charge is located above the Gaussian cylinder.
• Find the sign of the flux through:
Surface A: Surface B: Surface C:
• Can you tell whether the net flux through the Gaussian surface is positive, negative, or zero? Explain.
Learn your wayIncludes step-by-step video
Chapter 5 Solutions
Tutorials in Introductory Physics
Additional Science Textbook Solutions
University Physics Volume 2
College Physics
Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)
Conceptual Integrated Science
Conceptual Physics (12th Edition)
College Physics: A Strategic Approach (3rd Edition)
- C. The Gaussian cylinder encloses opposite charges of equal magnitude. (The charges are on the axis of the cylinder and equidistant from the center.) • Find the sign of the flux through: 40d Surface A: Surface B: Surface C: • Is the net flux through the Gaussian surface positive, negative, or zero?arrow_forwardWhich of the following is not a property of electric flux? O It is a vector quantity showing the path of the charge once it moves in an electric field. The answer cannot be found on the other choices. O It will neevr intersect with other flux lines. O It is an imaginary line showing the flow of the electric field.arrow_forwardTime left C For which of the following charge distributions would Gauss's law not be useful for calculating the electric field? a spherical shell of radius R with charge uniformly distributed over its surface O a. O b. an infinitely long cylinder of radius R with charge uniformly distributed over its surface Oc. a uniformly charged sphere of radius R an infinite planar sheet having constant surface charge density Oe a cylinder of radius R and height h with charge uniformly distributed over its surface NEXT PAGEarrow_forward
- c) a) sheet for formula number) 6= Φ = d) e) ON.m²/C Problem 1 • Sphere-A of radius RA = 24 cm and a total charge QA = -7x10-⁹ [C]. Sphere-B of radius RB = 6 cm and a total charge QB = 6x10-⁹ [c]. ● b) ₁, the net electric flux through the spherical surface of radius r = 12 cm from the center of the spheres is given by Оф1= Оф1= (6-10-⁹) 8.85-10-12 spheres. E2 = f) B ON.m²/C -790.96 g) spheres? E3 = The net electric flux can be calculated using the formula number (refer to the formula The figure shows two concentric conducting thin spherical shells: Calculate 1. RB RA [SI unit] (-1.10-⁹) 8.85-10-12 OP₁ = [SI unit] What is the SI unit of the electric flux? ON/C ON/(C.m²) Calculate E2, the magnitude the electric field at r = 35 cm from the center of the (-7-10-⁹) 8.85-10-12 [SI unit] What is the SI unit of the electric field? ON/(C.m²) ON/C Calculate E3, the magnitude of the electric field at r = 5 cm from the center of thearrow_forwardA thin wire is bent into the shape of the semicircle illustrated below. The radius of the semicircle is denoted by the symbol a and the negatively charged wire has total charge -Q and uniform charge density. Derive an equation for the electric field at the origin in terms of k, Q, a, and appropriate unit vectors. y dq Rarrow_forwardD. A positive charge is located above the Gaussian cylinder. • Find the sign of the flux through: Surface A: (1) Surface B: Surface C: • Can you tell whether the net flux through the Gaussian surface is positive, negative, or zero? Explain.arrow_forward
- A hollow metal sphere has 7 cm and 9 cm inner and outer radii, respectively, with a point charge at its center. The surface charge density on the inside surface is –250 nC/m² . The surface charge density on the exterior surface is +250 nC/m² . What is the strength of the electric field at point 12 cm from the center? Express your answer to three significant figures and include the appropriate units. • View Available Hint(s) HÀ ? a xa Xb b х-10п E Value N/C %3Darrow_forwardI Review The figures show cross sections of three-dimensional closed surfaces. They have a flat top and bottom surface above and below the plane of the page. However, the electric field is everywhere parallel to the page, so there is no flux through the top or bottom surface. The electric field is uniform over each face of the surface. Submit Previous Answers v Correct Part C For (Figure 3), does the surface enclose a net positive charge, a net negative charge, or no net charge? Match the words in the left column to the appropriate blanks in the sentences on the right. Rese Figure « 1 of 3 > an outward There is flux through each side of the surface, which indicates that there is an inward enclosed in this surface. zero no net charge 10 N/C 10 Ν/C a net negative charge 10 N/C 10 N/C a net positive charge Submit Request Answerarrow_forwardCharges q1 = -9Q and 92 +2Q are located at x = -a and x = +a, respectively. Part A What is the net electric flux through a sphere of radius 2a centered at the origin? Give your answer as a multiple of Q/e. Vα ΑΣφ ? Submit Request Answer Part B What is the net electric flux through a sphere of radius 2a centered at x = 2a? Give your answer as a multiple of Q/e0•arrow_forward
- Good morning could you help me to solve the following problem?Thanks in advanceA ring of radius a carries a uniformly distributed positive total charge. uniformly distributed. Calculate the electric field due to the ring at a point P which is at a distance x from its center, along the central axis perpendicular to the plane of the ring. Use fig. a The fig.b shows the electric field contributionsof two segments on opposite sides of the ring.arrow_forwardA system of very long conductors (Total length l) consists of an inner cylinder of radius a and a thin outer cylinder of radius b. The inner cylinder is given a total charge q while the outer cylinder has the same amount of charge but with the opposite sign. All charges are evenly distributed on the surface of the conductor. Determine: *The electric field throughout space!arrow_forwardProblem An electric charge Q is distributed uniformly along a thick and enormously long conducting wire with radius R and length L. Using Gauss's law, what is the electric field at distance r perpendicular to the wire? (Consider the cases inside and outside the wire) Solution To find the electric field inside at r distance from the wire we will use the Gauss's law which is expressed as DE - dÃ= Q piR2L We will choose a symmetric Gaussian surface, which is the surface a cylinder excluding its ends, then evaluate the dot product to obtain E A = qenc / epsilon0 (Equation 1) Case 1: Inside the wire Since, r falls inside the wire, then all the enclosed charge must be: qenc = Qr2/R2 On the other hand, the Gaussian surface inside the wire is given by A = 2pirL Using Equation 1, the electric field in simplified form is E = Qr / 2piR2Lepsilono Case 2: Outside the wire Since, r falls outside the wire, then, all the charge must be enclosed, thus qenc = On the other hand, the Gaussian surface…arrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning