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Ch 22: Gauss' Law

Chapter 22, Problem 22

A hemispherical surface with radius r in a region of uniform electric field E→ has its axis aligned parallel to the direction of the field. Calculate the flux through the surface.

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Hey, everyone today, we're dealing with the problem about flux. Now, we're being told that a curve lending is placed in a uniform electric field. E it's a uniform electric field. The radius of a flat circular area on the left of the lenses are the optical lens axis is parallel to the fuel's direction shown as shown in the image below. With this were being asked to find out the electrical flux through the curved surface. Now, while this may seem a little complicated, it's actually really quite easy to understand. So electrical flux, electrical flux can be expressed as the man through the electric field multiplied by the area through which the field is passing through. Now, the flux passes through a flat circle, the left part of this lens, right, which means the area of that circle will simply be E in two pi R squared. Now, so this is the area entering the curved or the flux entering the circular surface. However, every electric fuel line that passes through the flat side of the circle must also pass through the curved surface of the lens as shown here below. So the flux of the curved part of the lens will be the same as the flux of the flat. So flux curved must equal flux passing through the flat side, which means the flux is simply E pi R squared. The flux through the curved surface is e multiplied by pi R squared or answer choice. See, I hope this helps and I look forward to seeing you all in the next one.
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A hollow, conducting sphere with an outer radius of 0.250 m and an inner radius of 0.200 m has a uniform surface charge density of +6.37×10−6 C/m2. A charge of −0.500 μC is now introduced at the center of the cavity inside the sphere. (c) What is the electric flux through a spherical surface just inside the inner surface of the sphere?
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Textbook Question
A flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60° to a uniform electric field of magnitude 14 N/C. (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet? Why or why not?
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Textbook Question
A flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60° to a uniform electric field of magnitude 14 N/C. (c) For what angle φ between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest? Explain your answers.
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Textbook Question
You measure an electric field of 1.25×106 N/C at a distance of 0.150 m from a point charge. There is no other source of electric field in the region other than this point charge. (a) What is the electric flux through the surface of a sphere that has this charge at its center and that has radius 0.150 m? (b) What is the magnitude of this charge?
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Textbook Question
Charge q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00 cm from the center of the sphere, the electric field due to the charge distribution has magnitude E = 940 N/C. What are (a) the volume charge density for the sphere?
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Textbook Question
Charge q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00 cm from the center of the sphere, the electric field due to the charge distribution has magnitude E = 940 N/C. What are (b) the electric field at a distance of 2.00 cm from the sphere's center?
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