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.
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- Part A A2.0 cm × 3.0 cm rectangle lies in the xz-plane. What is the magnitude of the electric flux through the rectangle if E = (120î – 250k) N/C? Express your answer in newton meters squared per coulomb. • View Available Hint(s) de = 0 N. m2 /C Submit Previous Answers v Correct Part B What is the magnitude of the electric flux through the rectangle if E = (120î – 250ĵ) N/C? Express your answer in newton meters squared per coulomb. • View Available Hint(s) ΑΣφ. 圈] ? de = - 1.5 N-m² /CA. The Gaussian cylinder is in a uniform electric field of magnitude E, aligned with the cylinder axis. • Find the sign and magnitude of the flux through: Surface A: Surface B: Surface C: • Is the net flux through the Gaussian surface positive, negative, or zero?Problem 1 The figure shows two concentric conducting thin spherical shells: • Sphere-A of radius R = 7 cm and a total charge QA = -5×10-⁹ [C]. Sphere-B of radius RB = 27 cm and a total charge QB = 5×10-⁹ [C]. A $1 a) The net electric flux can be calculated using the formula number (refer to the formula sheet for formula number) b) ₁, the net electric flux through the spherical surface of radius r = 13 cm from the center of the spheres is given by O c) $1 d) = B = E2 f) (5.10-⁹) 8.85-10-12 ON.m²/C = $1 Calculate $₁. [SI unit] What is the SI unit of the electric flux? ON/C ON/(C.m²) e) Calculate E2, the magnitude of the electric field at r = 42 cm from the center of the spheres. RB ON/(C.m²) RA = [SI unit] (-5-10-⁹) 8.85-10-12 O Φ [SI unit] What is the SI unit of the electric field? ON.m²/C ON/C g) Calculate E3, the magnitude of the electric field at r = 6 cm from the center of the spheres? E3 = = (0-10-⁹) 8.85-10-12
- 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 theII Review | Constants A point charge of -3.00 µC is located in the center of a spherical cavity of radius 6.80 cm inside an insulating spherical charged solid. The charge density in the solid is 7.35 x 10 4 C/m³. Part A Calculate the magnitude of the electric field inside the solid at a distance of 9.30 cm from the center of the cavity. Express your answer with the appropriate units. HA E = Value Units Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining Part B Find the direction of this electric field. O radially outward O radially inwardA. Four closed surfaces, S₁ through S 4, together with the charges - 2Q, Q , and -Q are sketched in the Figure below. (The colored lines are the intersections of the surfaces with the page.) Find the electric flux through each surface. S₁ -20 SA B •+Q -2. S₂ Sz
- 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 %3DThe following charges are located inside a submarine: 6.20 μC, -9.00 μC, 27.0 µC, and -78 µC. (a) Calculate the net electric flux through the hull of the submarine. -1.36 What is the relationship between the total flux through a closed surface and the net charge within that surface? N. • m²/c (b) Is the number of electric field lines leaving the submarine greater than, equal to, or less than the number entering it? greater than equal to less thanExpress answer/s in mks (SI) . Draw diagram whenever necessary The electric field intensity 2m from the surface of a charged sphere is 30.86 x103 N/C. The charge densityat the surface of the sphere is 8.85 x 19‒7 coul/m2. What is the radius of the sphere ?• Hint: The Gaussian surface area of the big sphere of radius, r + 2 is : A = 4π (r +2 )2 Ans. r = 2.5 m
- Constants A point charge g = 4.05 nC is located on the x-axis at z = 2.20 m, and a second point charge q2 = -5.70 nC is on the y-axis at y = 1.00 m. Part A What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius ri = 0.325 m ? N.m?/C Submit Request Answer • Part B What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius r2 = 1.50 m ? Vo AEO ? = N.m?/C Submit Request Answer Part C What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius r3 = 2.85 m ? Πνα ΑΣφ = N- m² /C Submit Request AnswerRegion I: (outside the shell, r > b) • What is the formula for the area of the Gaussian surface? A = • What is the formula for the charge inside? Qin • What is the formula for the electric field? E = Region II: (within the shell, b>r> a) • What is the formula for the area of the Gaussian surface? A = • What is the formula for the charge inside? Qin = • What is the formula for the electric field? E = Region III: (inside the hole, r < a) • What is the formula for the electric field? E= NOTE: • Use & in your result. • Do not use k in your result. • The volume of a spherical shell is π [(router)³ - (inner)³] 3Problem 3. with their central axes along the x-axis. • Ring 1 has a radius R₁ = 3 cm and a charge Q₁ = 46 PC uniformly distributed over the ring. • Ring 2 has a radius R₂ = 2 cm and linear charge density X2 = 41 pc/m. • The rings are separated by a distance d = 30 cm. R₁ Ring 1 OE₁ = OE₁ = 5. y [cm] 4. point A is: Ē₁ - The figure shows two rings 1. The magnitude of the electric field created by ring 1 at point A can be calculated by: OE₁ = Coulombs. Q₂ = OE₂ = R₁ OE₂ = -- A (9x109) (46-10-12)-(3-10-2) ((3-10-2)²+(3-10-2)2) 1.5 2. The value of the magnitude of the electric field created by ring 1 at point A is: E₁ = [N/C] 3. field created by ring 1 at point A? ORight OLeft Oup Odown (9x109) (46-10-12)-(3-10-2) ((3-10-2)²+(3-10-2)2)0.5 (9x109) (46-10-12). (3) ((3)²+(3)2)1.5 d + What is the direction of the electric Ring 2 R₂2 The electric field created by ring 1 at [C] [N/C] Calculate the total charge of ring 2 in 6. The magnitude of the electric field created by ring 2 at point A can be…