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This is chapter 22 problem 6 from the sears and zamanski's university physics textbook.
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The cube in figure e 22 .6 has sides of length l equals 10 .0 centimeters.
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The electric field is uniform with magnitude e equals 4 .00 times 10 cube newton's per coulum and is parallel to the xy plane at an angle of 53 .1 degrees measured from the positive x axis toward the positive y axis.
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What is the electric flux to each of the six cube faces, s1, s2, s3, s4, s5, and s6? and what is the total electric flux through all faces of the cube? all right, so first thing i've done is i've sketched out the cube as shown in the figure.
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S1 is the left side.
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S2 is the top.
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S3 is this right side.
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S4 is the bottom.
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Of the cube.
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S5 is the front of the cube and then s6 which is the hardest one to represent is the very back of the cube.
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I've also drawn the electric field is not shown in the diagram but i've shown that in blue.
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The electric field is parallel to the x, y, plane as shown.
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So we imagine it's going parallel to the plane formed by the x and y axes, but it's at an angle of 53 .1 degrees measured from the positive x -axis going toward the positive y axis.
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So we can see that angle, which i've shown as theta, as 53 .1 degrees.
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So that's the angle as the electric field is going, stemming out from the x -axis.
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So it's uniform to the cube.
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So i've just sketched a couple of the electric field lines for illustrative purposes.
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So our basic definition for electric flux is what we're going to use for all six sides.
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The electric flux is simply the electric field multiplied by the area of each side times the cosine of this angle phi.
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And phi is the angle that is between the normal to the surface and the direction of the electric field.
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So we need to just figure out what phi is because that's what's going to change for each of the sides.
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So with this basic definition, well, we know the electric field in each case, it's uniform, 4 .00 times 10 cube newtons per coulum.
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And we know the area of each side.
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It's a cube, so it's just length times length.
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So the area is l squared, allowing us to rewrite the electric flux as el squared times the cosine of five.
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So we just need to repeat this calculation right here for each and every one of the sides.
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I'm not going to write this out for each and everyone because it's just plugging in.
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E, the electric field is always this value that we're given, and l is always 10 centimeters.
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And make sure to convert to meters from 10 centimeters each time that you are computing this calculation right here.
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So we always want to make sure we're in standard units.
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So we will just rewrite this basic one here.
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As that's the same thing as 0 .10 meters...