A uniform electric field of magnitude 2.80-104 N/C makes an angle of 47" with a plane surface of area 1.53×102 m² Part A What is the electric flux through this surface?
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![A uniform electric field of magnitude 2.80-10
N/C makes an angle of 47" with a plane surface
of area 1.53×102 m²
Part A
What is the electric flux through this surface?
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- A 39.0-cm-diameter circular loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be 5.42 x 105 Nm²/C. What is the magnitude of the electric field? MN/C Need Help? Read It Master itPart A A conducting sphere has a net charge of Q = 8nC and a radius of r= 10cm. What is the surface charge density? να ΑΣφ nC/m2 Submit Request AnswerPart A A closed surface encloses a net charge of 3.10 μC . What is the net electric flux through the surface? Express your answer in newton-meters squared per coulomb. %0 ΑΣΦ Submit Request Answer Part B Submit ? If the electric flux through a closed surface is determined to be 2.20 Nm²/C, how much charge is enclosed by the surface? Express your answer in coulombs. IVE ΑΣΦ Request Answer N.m²/C ? C Pearson
- I Rew A spherically symmetric charge distribution produces the electric field E =( 5700 r2)r N/C, where r is in m. Part A What is the electric field strength at r = 19.0 cm ? ? N/C Submit Request Answer Part B What is the electric flux through a 38.0-cm-diameter spherical surface that is concentric with the charge distribution? Nm2/C Submit Request Answer Part C How much charge is inside this 38.0-cm-diameter spherical surface? CA hollow sphere of radius 1.54 m is in a region where the electric field is radial and directed toward the center of the sphere. If the magnitude of the field at the surface of the sphere is 21.6 N/C, what is the net electric flux through the spherical surface? N. m²/cFind the electric flux through the closed surface whose cross-section is shown below. a a -2.0 × 10-6 C a. -2.00 × 10-N·m²/C O b. -2.00 × 10-6a² N·m²/C -2.25 × 105 N·m²/C c. d. -2.25 × 105a² N·m²/C
- A closed surface is in the form of a block with length c, width b and height a. In this room there is an electric field parallel to the x axis with magnitude: E = A + Bx, where A and B are constants. a. Determine the total flux over the closed surface. b. Determine the amount of charge covered by the payload. N x = a ko-The cube in the figure (Figure 1)has sides of length L=10.0 cm. The electric field is uniform, has a = magnitude E 4.00 x 103 N/C, and is parallel to the xy-plane at an angle of 36.9° measured from the axis toward the +y-axis. +2 What is the electric flux through the cube face S₁? 1 ΑΣΦ Φι Figure Submit Request Answer ▾ Part B What is the electric flux through the cube face S2? ΜΕ ΑΣΦ 42 = Submit Request Answer ▾ Part C What is the electric flux through the cube face S3? ΜΕ ΑΣΦ 43 = Submit Request Answer ▾ Part D 1 of 1 S₂ (top) S6 (back) What is the electric flux through the cube face S4? ΜΕ ΑΣΦ ? N.m²/C ? N-m²/C ? N. m²/C ? N.m²/CYou may want to review (Pages 673-676). Figure q (inside) (+) Q 1 of 1 Part A What is the net electric flux through the torus (i.e., doughnut shape) of the figure (Figure 1)? Assume that Q = 250nC and q = -3.0nC. Þ= IV- ΑΣΦ Submit Provide Feedback Request Answer F ? N m²/C