QuestionFind the net electric flux from a cube with edge $6.6 \mathrm{~m}$ in the field $\vec{E}=2 y \hat{j}$ N/C. Options: a. 1030, b. 575, c. 862, d. 287, e. 460 N.m²/C.

Studdy Solution

STEP 1

Assumptions1. The electric field is given by $\vec{}= y \hat{j}$ N/C. . The cube has an edge length of $6.6 \mathrm{~m}$ .
3. The cube is placed with one corner at the origin.
4. The net flux is calculated by integrating the electric field over the surface of the cube.

STEP 2

The electric field is only in the $\hat{j}$ direction, so it will only pass through the top and bottom faces of the cube. The other faces are perpendicular to the electric field, so the field does not pass through them.

STEP 3

The flux through a surface is given by $\Phi = \int \vec{} \cdot d\vec{A}$ , where $\vec{}$ is the electric field and $d\vec{A}$ is the differential area vector. The dot product means that only the component of the electric field that is perpendicular to the surface contributes to the flux.

STEP 4

For the top face of the cube, the area vector $d\vec{A}$ points in the positive $\hat{j}$ direction, the same as the electric field. So, the flux through the top face is $\Phi_{top} = \int \vec{} \cdot d\vec{A}$ .

STEP 5

The electric field $\vec{}=2 y \hat{j}$ depends on $y$ , the height above the origin. On the top face of the cube, $y$ varies from $0$ to $. \mathrm{~m}$ . So, we need to integrate over this range.

STEP 6

The differential area vector on the top face is $d\vec{A} = dx\,dz\,\hat{j}$ , where $dx$ and $dz$ are the differential lengths in the $x$ and $z$ directions, respectively. The total area of the top face is $(6.6 \mathrm{~m})^2$ .

STEP 7

Substitute the expressions for $\vec{}$ and $d\vec{A}$ into the integral for $\Phi_{top}$ and perform the integration.
$\Phi_{top} = \int0^{6.6}2 y\, dx\,dz =2 \int0^{6.6} y\, dx\,dz$

STEP 8

The integral of $y$ from $0$ to $6.6 \mathrm{~m}$ is $(1/2) y^2$ evaluated from $0$ to $6.6 \mathrm{~m}$ , which gives $(1/2) (6.6 \mathrm{~m})^2$ .

STEP 9

Multiply this by the area of the top face to get the total flux through the top face.
$\Phi_{top} =2 \times (/2) (6.6 \mathrm{~m})^2 \times (6.6 \mathrm{~m})^2 =287.496 \mathrm{~} \cdot \mathrm{m}2 / \mathrm{C}$

STEP 10

For the bottom face of the cube, the area vector $d\vec{A}$ points in the negative $\hat{j}$ direction, opposite to the electric field. So, the flux through the bottom face is $\Phi_{bottom} = -\int \vec{} \cdot d\vec{A}$ .

STEP 11

On the bottom face of the cube, $y$ is always $0$ , so the electric field is $0$ . Therefore, the flux through the bottom face is $0$ .

STEP 12

The total flux through the cube is the sum of the fluxes through the top and bottom faces.
$\Phi_{total} = \Phi_{top} + \Phi_{bottom}$

STEP 13

Substitute the values for $\Phi_{top}$ and $\Phi_{bottom}$ to calculate $\Phi_{total}$ .
$\Phi_{total} =287.496 \mathrm{~} \cdot \mathrm{m}2 / \mathrm{C} +0 =287.496 \mathrm{~} \cdot \mathrm{m}2 / \mathrm{C}$ The net flux from the cube is approximately $287 \mathrm{~} \cdot \mathrm{m}2 / \mathrm{C}$ , which corresponds to option d.

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