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Ch 22: Gauss' Law

Chapter 22, Problem 22

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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Everyone in this problem, we are asked to determine the values of the angle phi for which the electric flux magnitude through the surface is going to be first maximum and second minimum where we have a plane surface of area 50 millimeters squared in a uniform electric field of magnitude 25 newton per Coolum. The angle between the electric field and the normal to the surface is the one that's going to be the fight. And we want to determine the values of that. So the way you want to solve this problem is to first by recalling that the electric flux flux for a uniform electric field or flat surface is given by E. So five e here is going to be the electric flux for uniform electric field at a flat surface. And it can be calculated by multiplying E multiplied by a by coastline of the angle or the fi so E is the electric field. The A is the area of the plane surface and the fire here is the angle between the electric field and the normal to the surface. So this is the one that we want to find. This is the angle between the electric field and normal to the surface. So essentially, we want to find a five value that will actually first maximize this and second, minimize this okay. So for the first part is for maximum electric flux or five E, so the maximum five E is going to actually be equal to by a equals two e multiplied by a or essentially where or when Cosigned five equals to one. So we essentially just find one to find a five value where the cosine value of that is one. So therefore, we can actually find this by taking the in first off the fi. So remember that if I can be find here by taking the inverse of cost, which is going to be our cause of one Where the fi is then going to actually be 0°. Therefore, The fight that will maximize the value of the electric flux will have to be 0°. So the angle between the electric field and normal to the surface has to be 0°. Next, we want to find the minimum FI E or minimum electric flux. So five E will be minimum when it is actually zero and it will be zero when cosine sine of five also equals to zero. And we want to find the angle by taking the in. First of course, sandwiches are co co sign of zero, which will give us a five value of 90 degrees. Therefore five. Well, it goes to 90 degrees just like. So, so that will be all for this problem. So the fire value that will maximize the electric field or the electric flux is zero degrees and the five value that will minimize it is actually 90 degrees. So that will correspond to answer option B and that will be all for this particular practice problem. So if you guys have any sort of confusion, please make sure our other less to watch our other lesson videos or on similar topics and that will be all for this video. Thank you.
Related Practice
Textbook Question
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. (b) Calculate the strength of the electric field just outside the sphere?
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Textbook Question
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 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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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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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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