To calculate the electric flux through a surface of arbitrary shape you need to: a. multiply...
To calculate the electric flux through a surface of arbitrary shape you need to:
a. multiply the total area of the surface by the average magnitude of the electric field at the surface.
b. split the surface into small areas, and then add up the product of the perpendicular component of the electric field through each small area times the small area.
c. split the surface into small areas, and then add up the product of the electric field times the small area.
Homework Answers
Flux=E.ds,
Where E is the electric field and ds is the small area and we take their dot product which means that it is the product of the perpendicular component of electric field and the small area. Integrating it simply means adding them up. So option b is the correct option
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To calculate the electric flux through a surface of arbitrary
shape you need to:
a. multiply...
Learning Goal: To understand the definition of electric flux, and how to calculate it. Flux is...
Learning Goal: To understand the definition of electric flux, and how to calculate it. Flux is the amount of a vector field that "flows" through a surface. We now discuss the electric flux through a surface (a quantity needed in Gauss's law): ΦE=∫E⃗ ⋅dA⃗ , where ΦE is the flux through a surface with differential area element dA⃗ , and E⃗ is the electric field in which the surface lies. There are several important points to consider in this expression:...
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?
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?
oy direction 13. The electric flux through a surface: A) is the amount of electric field...
oy direction 13. The electric flux through a surface: A) is the amount of electric field piercing the surface B) is the electric field multiplied by the area. C) does not depend on the area involved. D) is the line integral of the electric field around the edge of the surface. E) is the amount of electric field skimming along the surface.
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.
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Electric charge produces an electric field, and the flux of that field passing through any closed surface is proportional to the total charge contained within that surface. In other words, if you have a real or imaginary closed surface of any size and shape and there is no charge inside the surface, the electric flux through the surface must be zero. (i) In your own understanding kindly represent the above statement to its mathematical formula.
A flat surface of area 2.60 m2 is rotated in a uniform electric field of magnitude...
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oy direction 13. The electric flux through a surface: A) is the amount of electric field piercing the surface B) is the electric field multiplied by the area. C) does not depend on the area involved. D) is the line integral of the electric field around the edge of the surface. E) is the amount of electric field skimming along 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...
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