The cube in fig 23 31 has edge length 140 mathrmm and is...

The cube in Fig. $23-31$ has edge length 1.40 $\mathrm{m}$ and is oriented as shown in a region of uniform electric field. Find the electric flux through the right face if the electric field, in newtons per coulomb, is given by (a) 6.00 $\mathrm{i}$ , (b) $-2.00 \hat{\mathrm{j}},$ and $(\mathrm{c})-3.00 \hat{\mathrm{i}}+4.00 \hat{\mathrm{k}}$ . (d) What is the total flux through the cube for each field?

Transcript

00:01 Okay, so for this problem, you're given the few different electric fields and a cube, and you're asked to compute the flux through the right face.

00:18 So that's this one.

00:22 There's our right face.

00:24 So before i start, so let's just identify what is flux.

00:29 So flux is the integral of the electric field with respect to.

00:34 The area.

00:36 And so the electric field is the same everywhere here in each of the cases.

00:41 In each of the cases, the electric fields are spatially constant, like they're just a constant number.

00:47 And so this always is just going to simplify to the dot product of eda.

00:55 Oh, sorry, not da.

00:58 E.

01:00 Dot a.

01:02 Wonderful.

01:02 And so the electric field changes in each case, but the area vector does not.

01:09 So i'm just drawing the right face.

01:12 So that's like this right face and identifying the area vector.

01:17 So the area vector always points normal to the surface, so perpendicular to the surface, to identify it.

01:26 And since, so let's now identify its sort of mathematical character, or what is the actual vector mathematically? a points along the y axis.

01:40 And so you can say a is like whatever its magnitude, so the magnitude of the area times the j hat vector unit vector direction.

01:53 And so the area is just the length times the length.

01:59 So 1 .4 meters squared and then that's along the j hat direction.

02:06 So 1 .4 meters squared is 1 .96 meters square along the j hat direction.

02:13 So we have the a identify and i'm just going to kind of put it over here.

02:18 And now we can go through each case and ask what is the dot product between e and a.

02:24 So for a, the electric field is, i'm just going to double check.

02:30 It is 6 i -hat so that means the electric field is just pointing along the i -hat direction and so because by the rules of dot product if two vectors are perpendicular i .e.

02:49 Entirely along different directions then their dot product is zero so you can say e .a is zero.

03:01 Okay and for b, the electric field is, i could have to keep looking back because, you know, you can kind of ignore certain pieces of information.

03:19 So let's see, that's minus 2 j hat.

03:22 Okay, so it's just in the j hat direction...

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