A uniform electric field ai + bj intersects a surface of area A. What is the flux through this area if the surface lies in the yz plane
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A: Let E be defined as the electric field, and A be defined as area. Then flux be given as, ϕ=E.A
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A: Given: The electric flux through a sphere is ϕ1=10.0 Nm2/C,
Q: Don't use chatgpt.
A: Step 1:Given:a=1.4mb=3.6mE=8.9V/mθ=63.5∘Unknown:The electric flux through the annulus: ϕE=?Step 2:…
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Q: Normal
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Q: The square surface shown in the figure measures 2.38 cm on each side. It is immersed in a uniform…
A: Here is the step-by-step explanation:Stepe 1: Use formula:We will use the formula, to calculate the…
Q: The figure shows a circular disk of radius 54 mm in a uniform electric field of magnitude 5.56 x 103…
A: We will first write an expression for electric flux. Then we apply it to the given situation. The…
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A: Given Length l = 0.400 m breadth b = 0.600 m Electric field E = 95.0 N/C Angle θ…
Q: The figure below shows a closed cylinder with cross-sectional area A 3.20 m². The constant electric…
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Q: A cube of side L = 2.2 m lies in a region where the electric field is given by 3.G take → 3.6 en…
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Q: Consider a triangle in the presence of a uniform electric field given by 5.6 i N/C. The endpoints of…
A: Please see the answer below.
Q: The figure shows a Gaussian surface in the shape of a cube with edge length 1.70 m. What are (a) the…
A: Edge length = 1.70 mE = 4.50y N/C
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- The figure below shows a uniform electric field of magnitude E = 420 N/C making an angle of o = 66.5° with a flat surface of area A = 3.20 m2. What is the magnitude of the electric flux through this surface (in N · m2/C)? E N. m2/cA uniform electric field of magnitude 5.7 x 104 N/C is at an angle of 10° to a square sheet with sides 5.5 m long. What is the electric flux through the sheet? Hint Electric flux is E N.m²/C.A uniform electric field of magnitude 3.6×104N/C is at an angle of 80° to a square sheet with sides 8.5 m long. What is the electric flux through the sheet?
- An electric flux of 18.35 Nm2/C passes through a flat surface that is perpendicular to a constant electric field of strength 11.5 N/C. What is the area of the surface?A hollow sphere of radius .05m contains a negative charge sitting right at its center. (The sphere itself is not charged.) The electric field everywhere on the surface of the sphere has the magnitude 3.0*10^-4 N/C. Draw the sphere and show the electric field vectors on your drawing. What is the total electric flux through the surface of the sphere?A point charge is located at the origin. Centered along the x axis is a cylindrical closed surface of radius 10 cm with one end surface located at x = 2 m and the other end surface located at x = 2.5 m. If the magnitude of the electric flux through the surface at x = 2 m is 4 N . m2 /C, what is the magnitude of the electric flux through the surface at x = 2.5 m? Select one: a. 1.8 N . m2 /C b. 2.56 N . m2 /C c. 1.0 N . m2 /C d. 4.0 N . m2 /C e. 5.0 N . m2 /C
- A uniform electric field with a magnitude of 1.25x105 N/C passes through a rectangle with sides of 2.50 m and 5.0 m . If the electric flux through the rectangle is 6.6x105 Nm2/C , the vector normal to the rectangular surface ?The square surface shown in the figure measures 3.4 mm on each side. It is immersed in a uniform electric field with magnitude E = 1400 N/C and with field lines at an angle of 35° with a normal to the surface, as shown. Take that normal to be "outward," as though the surface were one face of a box. Calculate the electric flux through the surface. Normal 4 Number i Units