Course Contents » ... » Homework 03 » Flux Above Surface An electric flux of 159...
Course Contents » ... » Homework 03 » Flux Above Surface An electric flux of 159 N.m2/C passes through a flat horizontal surface that has an area of 0.77 m2. The flux is due to a uniform electric field. What is the magnitude of the electric field if the field points 15o above the horizontal?
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Course Contents » ... » Homework 03 » Flux Above Surface An electric flux of 159...
An electric flux of 143 N.m2/C passes through a flat horizontal surface that has an area...
An electric flux of 143 N.m2/C passes through a flat horizontal surface that has an area of 0.62 m2. The flux is due to a uniform electric field. What is the magnitude of the electric field if the field points 15o above the horizontal?
An electric flux of 157 N·m2/C passes through a flat horizontal surface that has an area...
An electric flux of 157 N·m2/C passes through a flat horizontal surface that has an area of 0.72 m2. The flux is due to a uniform electric field. What is the magnitude of the electric field if the field points 15° above the horizontal? Submit Answer Tries 0/5
A flat surface of area 3.10 m2 is rotated in a uniform electric field of magnitude...
A flat surface of area 3.10 m2 is rotated in a
uniform electric field of magnitude E = 5.20 105 N/C.
(a) Determine the electric flux through this area when the
electric field is perpendicular to the surface.
N · m2/C
(b) Determine the electric flux through this area when the electric
field is parallel to the surface.
N · m2/C
A flat surface of area 2.60 m2 is rotated in a uniform electric field of magnitude...
A flat surface of area 2.60 m2 is rotated in a uniform electric field of magnitude E-5.90 x 105 N/C. (a) Determine the electric flux through this area when the electric field is perpendicular to the surface. The correct answer is not zero. N m2/C (b) Determine the electric flux through this area when the electric field is parallel to the surface. 15 34 Your response differs significantly from the correct answer. Rework your solution from the beginning and check...
A collection of 11 electrons and 4 protons are contained in a Gaussian (i.e., closed) surface....
A collection of 11 electrons and 4 protons are contained in a Gaussian (i.e., closed) surface. What is the total electric flux through the surface? 15 eleo 7 eleo 7 eleo 4 eleo Suppose a square surface of side 0.10 m is placed in a uniform electric field of magnitude 50.0 N/C. If the angle between the field and the normal to the surface is 25.0°, what is the magnitude of the electric flux through the surface? 4.5 N-m2/ 0.45...
Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.46 T. When...
Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.46 T. When viewed edge-on, the first surface is tilted at an angle of 14° from the horizontal, and a net magnetic flux of 8.50 x 10-3 Wb passes through it. The same net magnetic flux passes through the second surface. (a) Determine the area of the first surface. m2 (b) Find the smallest possible value for the area of the second surface. m2
A flat surface with area of 7.00 m2 is oriented at an angle q to a...
A flat surface with area of 7.00 m2 is oriented at an angle q to a uniform electric field with magnitude 6.0 × 105 N/C. If the flux through this surface is 2.0 × 106 N×m2/C, what is q? 81.8° 32.9° 6.18° 61.6°
(a) The electric field in a certain region is E (4.0)) N/C. Determine the electric flux...
(a) The electric field in a certain region is E (4.0)) N/C. Determine the electric flux due to this freld through an area represented by the vector A (1.51-7.k) m 29 How is the flux defined in terms of the electric field vector, and the area vector? Review dot product rules. N·m2/C (b) Determine the flux due to the same electric field when the surface orientation has changed such that the area is now represented by the vector A -(1.51-7.3)...
9 -2 points SerPSET9 24 P001 Aflat surface of area 2.60 m2 is rotated in a...
9 -2 points SerPSET9 24 P001 Aflat surface of area 2.60 m2 is rotated in a uniform electric field of magnitude E = 5.90 x 10s NC (a) Determine the electric flux through this area when the electric field is perpendicular to the surface (b) Determine the electric flux through this area when the electric field is parallel to the surface.
An electric flux of 157 N·m2/C passes through a flat horizontal surface that has an area of 0.72 m2. The flux is due to a uniform electric field. What is the magnitude of the electric field if the field points 15° above the horizontal? Submit Answer Tries 0/5
A flat surface of area 3.10 m2 is rotated in a
uniform electric field of magnitude E = 5.20 105 N/C.
(a) Determine the electric flux through this area when the
electric field is perpendicular to the surface.
N · m2/C
(b) Determine the electric flux through this area when the electric
field is parallel to the surface.
N · m2/C
A flat surface of area 2.60 m2 is rotated in a uniform electric field of magnitude E-5.90 x 105 N/C. (a) Determine the electric flux through this area when the electric field is perpendicular to the surface. The correct answer is not zero. N m2/C (b) Determine the electric flux through this area when the electric field is parallel to the surface. 15 34 Your response differs significantly from the correct answer. Rework your solution from the beginning and check...
A collection of 11 electrons and 4 protons are contained in a Gaussian (i.e., closed) surface. What is the total electric flux through the surface? 15 eleo 7 eleo 7 eleo 4 eleo Suppose a square surface of side 0.10 m is placed in a uniform electric field of magnitude 50.0 N/C. If the angle between the field and the normal to the surface is 25.0°, what is the magnitude of the electric flux through the surface? 4.5 N-m2/ 0.45...
Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.46 T. When viewed edge-on, the first surface is tilted at an angle of 14° from the horizontal, and a net magnetic flux of 8.50 x 10-3 Wb passes through it. The same net magnetic flux passes through the second surface. (a) Determine the area of the first surface. m2 (b) Find the smallest possible value for the area of the second surface. m2
(a) The electric field in a certain region is E (4.0)) N/C. Determine the electric flux due to this freld through an area represented by the vector A (1.51-7.k) m 29 How is the flux defined in terms of the electric field vector, and the area vector? Review dot product rules. N·m2/C (b) Determine the flux due to the same electric field when the surface orientation has changed such that the area is now represented by the vector A -(1.51-7.3)...
9 -2 points SerPSET9 24 P001 Aflat surface of area 2.60 m2 is rotated in a uniform electric field of magnitude E = 5.90 x 10s NC (a) Determine the electric flux through this area when the electric field is perpendicular to the surface (b) Determine the electric flux through this area when the electric field is parallel to the surface.
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